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THE THIRTEEN BOOKS OF 
EUCLID'S ELEMENTS 

TRANSLATED FROM THE TEXT OF HEIBERG 

WITH INTRODUCTION AND COMMENTARY 



BY 

Sir THOMAS L. HEATH, 

K.C.B., K.C.Y.O., F.R.S., 

SC.D. CAMB,, HON, D.SC. OXFORD 
HONORARY FELLOW (SOMETIME FELLOW) OF TRINITY COLLEGE CAMBRIDGE 



SECOND EDITION 
REVISED WITH ADDITIONS 

VOLUME III 
BOOKS X— XIII AND APPENDIX 



DOVER PUBLICATIONS, INC. 

NEW YORK 






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pany. Ltd., 30 Lesmill Road, Don Mills, Toronto, 
Ontario. 

Published in the United Kingdom by Constable 
and Company, Ltd,, 10 Orange Street, London 
WC 2. 



This Dover edition, first published in 1956. is an 
unabridged and unaltered republication of the 
second edition. It is published through special ar- 
rangement with Cambridge University Press, 



Library of Congress Catalog Card Number: 56-4336 

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CONTENTS OF VOLUME III. 

PACK 

Book X. Introductory note * 

Definitions 1 10 

Propositions i — 47 14-101 

Definitions ii 101 

Propositions 48 — 84 102-177 

Definitions hi 177 

Propositions 85 — 115 178-254 

Ancient extensions of theory of Book X . 255 

Book XI. Definitions 260 

Propositions 77* 

Book XII. Historical note 365 

Propositions 569 

Book XI I L Historical note 438 

Propositions 440 

Appendix. I. The so-called "Book XIV." (by Hypsicles) $h 
II. Note on the so-called "Book XV." . 519 

General Index: Greek 521 

„ „ English 527 









BOOK 


X. 





INTRODUCTORY NOTE. 

We have seen (Vol. r., p. 351 etc.) that the discovery of the irrational is 
due to the Pythagoreans. The first scholium on Book x. of the Element 1 
states that the Pythagoreans were the first to address themselves to the in- 
vestigation of commensurability, having discovered it by means of their obser- 
vation of numbers. They discovered, the scholium continues, that not all 
magnitudes have a common measure. " They called all magnitudes measure- 
able by the same measure commensurable, but those which are not subject to 
the same measure incommensurable, and again such of these as are measured 
by some other common measure commensurable with one another, and such 
as are not, incommensurable with the others. And thus by assuming tbeir 
measures they referred everything to different com mensurabili ties, but, though 
they were different, even so (they proved that) not all magnitudes are com- 
mensurable with any. (They showed that) all magnitudes can be rational 
(pp-a) and all irrational (SXaya) in a relative sense ((Js wpoi t(); hence the 
commensurable and the incommensurable would be for them natural (kinds) 
(i^iacrti), while the rational and irrational would rest on assumption or con- 
vention (9itrn)." The scholium quotes further the legend according to which 
" the first of the Pythagoreans who made public the investigation of these 
matters perished in a shipwreck," conjecturing that the authors of this story 
"perhaps spoke allegorical ly, hinting that everything irrational and formless 
is properly concealed, and, if any soul should rashly invade this region of life 
and lay it open, it would be carried away into the sea of becoming and be over- 
whelmed by its unresting currents." There would be a reason also for keeping 
the discovery of irrationals secret for the time in the fact that it rendered un- 
stable so much of the groundwork of geometry as the Pythagoreans had based 
upon the imperfect theory of proportions which applied only to numbers. We 
have already, after Tannery, referred to the probability that the discovery 
or incommensurability must have necessitated a great recasting of the whole 
fabric of elementary geometry, pending the discovery of the general theory 
of proportion applicable to incommensurable as well as to commensurable 
magnitudes. 

It seems certain that it was with reference to the length of the diagonal of 
a square or the hypotenuse of an isosceles right-angled triangle that the irra- 
tional was discovered. Plato (Theaetetus, 1470) tells us that Theodoras of 
Cyrene wrote about square roots (Sura^u;), proving that the square roots of 



a BOOK X 

three square feet and five square feet are not commensurable with that of one 
square foot, and so on, selecting each such square root up to that of 1 7 square 
feet, at which for some reason he stopped. No mention is here made of Ja, 
doubtless for the reason that its incommensurability had been proved before. 
Now we are told that Pythagoras invented a formula for finding right-angled 
triangles in rational numbers, and in connexion with this it was inevitable that 
the Pythagoreans should investigate the relations between sides and hypo- 
tenuse in other right-angled triangles. They would naturally give special 
attention to the isosceles right-angled triangle ; they would try to measure the 
diagonal, would arrive at successive approximations, in rational fractions, to 
the value of J 2, and would find that successive efforts to obtain an exact 
expression for it failed. It was however an enormous step to conclude that 
such exact expression was impossible, and it was this step which the Pytha- 
goreans made. We now know that the formation of the side- and diagonal- 
n umbers explained by Theon of Smyrna and others was Pythagorean, and 
also that the theorems of Eucl. R. 9, 10 were used by the Pythagoreans in 
direct connexion with this method of approximating to the value of Jx. The 
very method by which Euclid proves these propositions is itself"an indication 
of their connexion with the investigation of J 2, since he uses a figure made 
up of two isosceles right-angled triangles. 

The actual method by which the Pythagoreans proved the incommensura- 
bility of J2 with unity was no doubt that referred to by Aristotle (Anal, prior. 
1. 23, 41 a 26 — j), a redwtio ad aiisurdum by which it is proved that, if the 
diagonal is commensurable with the side, it will follow that the same number 
is both odd and even. The proof formerly appeared in the texts of Euclid as 
x. 117, but it is undoubtedly an interpolation, and August and Heiberg 
accordingly relegate it to an Appendix. It is in substance as follows. 

Suppose AC, the diagonal of a square, to be com men- a 

su ruble with AH, its side. Let a : j3 be their ratio expressed 
in the smallest numbers. 

Then a > fi and therefore necessarily > 1. 

Now AC* ; A£* = a' ; &, 

and, since AC = xAB*, [Eucl. 1. 47] 

a.' = 2p. 

Therefore a a is even, and therefore a is even. 

Since a : J3 is in its lowest terms, it follows that /3 must be odd. 

Put a = 2y ; 

therefore qf = 2fp, 

or 0> = 2v», 

so that $", and therefore /J, must be even. 

But was also odd : 
which is impossible.-. 

This proof only enables us to prove the incommensurability of the 
diagonal of a square with its side, or of ^2 with unity. In order to prove 
the incommensurability of the sides of squares, one of which has three times 
the area of another, an entirely different procedure is necessary ; and we find 
in fact that, even a century after Pythagoras' time, it was still necessary to use 
separate proofs (as the passage of the TAeaeietus shows that Theodorus did) 
to establish the incommensurability with unity of ^3, ^/ 5, ... up to ^17. 




INTRODUCTORY NOTE 3 

This fact indicates clearly that the general theorem in Eucl. x. 9 that squares 
which have not to one another the ratio of a square number to a square number 
have their sides incommensurable in length was not arrived at all at once, but 
was, in the manner of the time, developed out of the separate consideration 
of special cases (Hankel, p. 103). 

The proposition x. 9 of Euclid is definitely ascribed by the scholiast to 
Theaetetus. Theaetetus was a pupil of Theodorus, and it would seem clear 
that the theorem was not known to Theodorus. Moreover the Platonic 
passage itself {Theatt. 147 d sqq.) represents the young Theaetetus as striving 
after a general conception of what we call a surd. "The idea occurred to 
me, seeing that square roots (Jim/hk) appeared to be unlimited in multitude, 
to try to arrive at one collective term by which we could designate all these 
square roots — I divided number in general into two classes. The number 
which can be expressed as equal multiplied by equal {urw utokk) I likened 
to a square in form, and I called it square and equilateral... .The intermediate 
number, such as three, five, and any number which cannot be expressed as 
equal multiplied by equal, but is either less times more or more times less, so 
that it is always contained by a greater and less side, I likened to an oblong 

figure and called an oblong number Such straight lines then as square the 

equilateral and plane number I defined as length ((ijmt), and such as square 
the oblong square roots (Svi-a^cts), as not being commensurable with the 
others in length but only in the plane areas to which their squares are 
equal. " 

There is further evidence of the contributions of Theaetetus to the theory 
of incommensurables in a commentary on Eucl. x. discovered, in an Arabic 
translation, by Woepcke (Aft moires prlsentis d fAcadimie des Sciences, xiv., 
1856, pp. 658 — 720). It is certain that this commentary is of Greek origin. 
Woepcke conjectures that it was by Vettius Valens, an astronomer, apparently 
of Antioch, and a contemporary of Claudius Ptolemy (and cent. a.d.). 
Heiberg, with greater probability, thinks that we have here a fragment of the 
commentary of Pappus (Euklid-studien, pp. 169 — 71), and this is rendered 
practically certain by Suter (Die Afathemqtikcr und Astronomen der Araber 
und ihre Werke, pp. 49 and in). This commentary states that the theory 
of irrational magnitudes " had its origin in the school of Pythagoras. It was 
considerably developed by Theaetetus the Athenian, who gave proof, in this 
part of mathematics, as in others, of ability which has been justly admired. 
He was one of the most happily endowed of men, and gave himself up, with a 
fine enthusiasm, to the investigation of the truths contained in these sciences, 
as Plato bears witness for him in the work which he called after his name. As 
for the exact distinctions of the above-named magnitudes and the rigorous 
demonstrations of the propositions to which this theory gives rise, I believe 
that they were chiefly established by this mathematician; and, later, the 
great Apollonius, whose genius touched the highest point of excellence in 
mathematics, added to these discoveries a number of remarkable theories 
after many efforts and much labour. 

"For Theaetetus had distinguished square roots [puissances must be the 
Swdjiw! of the Platonic passage] commensurable in length from those which 
are incommensurable, and had divided the well-known species of irrational 
lines after the different means, assigning the medial to geometry, the binomial 
to arithmetic, and the apotomc to harmony, as is stated by Eudemus the 
Peripatetic. 

" As for Euclid, he set himself to give rigorous rules, which he established, 



4 BOOK X 

relative to commensurability and incommensurability in general; he made 
precise the definitions and the distinctions between rational and irrational 
magnitudes, he set out a great number of orders of irrational magnitudes, and 
finally he clearly showed their whole extent." 

The allusion in the last words must apparently be to X. 115, where it is 
proved that from the medial straight line an unlimited number of other 
irrationals can be derived, all different from it and from one another. 

The connexion between the media/ straight line and the geometric mean 

is obvious, because it is in fact the mean proportional between two rational 

straight lines "commensurable in square only." Since ^(x+y) is the arithmetic 

mean between x. y\ the reference to it of the binomial can be understood. 

The connexion between the apotome and the harmonic mean is explained by 

some propositions in the second book of the Arabic commentary. The 

2xy 
harmonic mean between x, y is - , and propositions of which Woepcke 

quotes the enunciations prove that, if a rational or a medial area has for one 
of its sides a binomial straight line, the other side will be an apotome of corre- 
sponding order (these propositions are generalised from Euci. x. 1 r 1 — 4) ; the 

fact is that -S~ = -~—. . (x- y). 
x+y x* — yr 

One other predecessor of Euclid appears to have written on irrationals, 
though we know no more of the work than its title as handed down by 
Diogenes Laertius 1 . According to this tradition, Democritus wrote iript 
aXoyiov ypa/ifiar xal raa-rwr /3', two Books on irrational straight lints and 
solids (or atoms). Hultsch (Neue Jahrbikhtr fur Phitologie vnd Padagogik, 
1SS1, pp. 578 — 9) conjectures that the true reading may be -atpX iXoymy 
ypa/ifium jcAmftw, " on irrational broken lines." Hultsch seems to have 
in mind straight lines divided into two parts one of which is rational 
and the other irrational (" Aus einer Art von Umkehr des Pythagoreischcn 
Lehrsatzes iiber das rechtwtnklige Dreieck gieng zun&chst mit Letchtigkeit 
hervor, dass man eine Linie construtren konne, welche als irrational zu 
bezeichnen ist, aber durch Brechung sich darstellen' lasst ais die Summe 
einer rationalen und einer irrationalen Linie"). But I doubt the use of kXsoto? 
in the sense of breaking one straight line into parts ; it should properly mean 
a bent line, i.e. two straight lines forming an angle or broken short off at their 
point of meeting. It is also to be observed that vamav is quoted as a 
Democritean word (opposite to xivbv) in a fragment of Aristotle (202). I see 
therefore no reason for questioning the correctness of the title of Democritus' 
book as above quoted*. 

I will here quote a valuable remark of Zeuthen's relating to the classifi- 
cation of irrationals. He says {Gtschithte der Mathematik im Altertum und 
Mittela/ter, p. 56) "Since such roots of equations of the second degree as are 
incommensurable with the given magnitudes cannot be expressed by means 
of the latter and of numbers, it is conceivable that the Greeks, in exact 
investigations, introduced no approximate values but worked on with the 
magnitudes they had found, which were represented by straight lines obtained 
by the construction corresponding to the solution of the equation. That is 
exactly the same thing which happens when we do not evaluate roots but content 
ourselves with expressing them by radical signs and other algebraical symbols. 
But, inasmuch as one straight line looks like another, the Greeks did not get 

1 Diog. Laert. ix. 47, p. 139 (ed. Cobet). 
* Cf. unit. Vol. i., p. 41 j. 



INTRODUCTORY NOTE 5 

the same clear view of what they denoted (Le. by simple inspection) as our 
system of symbols assures to us. For this reason it was necessary to under- 
take a classification of the irrational magnitudes which had been arrived at by 
successive solution of equations of the second degree." To much the same 
effect Tannery wrote in 1883 (De la solution giometrique des firoblemes du 
second degri avant Euclide in Aflmoirts de la Soci/li dts sciences physiques et 
naturilUs de Bordeaux, 2* Serie, iv. pp. 395 — 416). Accordingly Book x. 
formed a repository of results to which could be referred problems which 
depended on the solution of certain types of equations, quadratic and biquad- 
ratic but reducible to quadratics. 
Consider the quadratic equations 

x? ± iax . p ± 8 . p' = o, 

where p is a rational straight line, and a, B are coefficients. Our quadratic 
equations in algebra leave out the p ; but I put it in, because it has always to 
be remembered that Euclid's x is a straight line, not an algebraical quantity, 
and is therefore to be found in terms of, or in relation to, a certain assumed 
rational straight tint, and also because with Euclid p may be not only of the 

form a, where a represents a units of length, but also of the form */ — ■ <*> 

which represents a length "commensurable in square only" with the unit of 
length, or ^A where A represents a number {not square) of units of area. 
The use therefore of p in our equations makes it unnecessary to multiply 
different cases according to the relation of p to the unit of length, and has the 
further advantage that, e.g., the expression p ± Jk , p is just as general as the 
expression Jhp±J^-p, since p covers the form Jh . p, both expressions 
covering a length either commensurable in length, or "commensurable in 
square only," with the unit of length. 

Now the positive roots of the quadratic equations 
x? ± tax . p ± ff , p* = o 
can only have the following forms 

^pfa + Va^/S), j^p^-vV^S) I 
■ r i = P (vV + B + a), X,' = p (>Ja r +~$ -a) i ' 

The negative roots do not come in, since x must be a straight line. The 
omission however to bring in negative roots constitutes no loss of generality, 
since the Greeks would write the equation leading to negative roots in another 
form so as to make them positive, i.e. they would change the sign of x in the 
equation. 

Now the positive roots jc,, x,', x t , x 3 ' may be classified according to the 
character of the coefficents a, ft and their relation to one another. 

I. Suppose that a, B do not contain any surds, i.e. are either integers or 

of the form mjn, where m, n are integers. 

Now in the expressions for x„ x,' it may be that 

m* 
( 1 ) 8 is of the form -= a 1 . 
fr 

Euclid expresses this by saying that the square on op exceeds the square 

on pi/a 1 — B by the square on a straight line commensurable in length with ap. 

In this case x, is, in Euclid's terminology, a first binomial straight line, 

and x, a first afiotome. 



BOOK X 

m* 
(a) In general, B not being of the form -^ a*, 

x l is a. fourth binomial, 

Xi 3. fourth apotomt. 
Next, in the expressions for x„ x t ' it may be that 
(1) B is equal to -j (V + 0), where w, « are integers, i.e. B is of the form 



n*-m* 

Euclid expresses this by saying that the square on pv<t" + B exceeds the 
sq uare o n op by the square on a straight line commensurable in length with 
P-Jj + B. 

In this case x, is, in Euclid's terminology, a second binomial, 
x t ' a second apotomt. 

(2) In general, B not being of the form -j , a', 

x, is a fifth binomial, 
x t ' a. fifth apotome. 

II, Now suppose that a. is of the form »/ — , where m, n are integers, and 

let us denote it by ,/A. 
Then in this case 

x, = p(J*+J\-B), x{ = p{,J\-jJ>^~&), 

xt m P ( jx+p + jk), x,' = P (jx~+p~- ,/A). 

Thus x It x,' are of the same form as x t , x t '. 

If <JX - 8 in jt,, *,' is not surd but of the form mjn, and if *A + B in x,, x f ' 
is not surd but of the form mjn, the roots are comprised among the forms 
already shown, the first, second, fourth and fifth binomials and apotomes. 

If s/k- B in jc,, Xi is surd, then 

m* 

(1) we may ha-ve of the form -j A, and in this case 

x, is a rift/n/ binomial straight line, 
*i' a third apotomt; 

(2) in general, B not being of the form -3 A, 

x, is a Jtxr/ji binomial straight line, 
*l' a /£rM apotome. 

With the expressions for *„ *,' the distinction between the third and sixth 
binomials and apotomes is of course the distinction between the cases 

(i) in which 8 = — (X + B), or 8 is of the form -* s A, 

and <i) in which B is not of this form. 

If we take the square root of the product of p and each of the six 
binomials and six apotomes just classified, i.e. 



INTRODUCTORY NOTE 7 

in the six different forms that each may take, we find six new irrationals with 
a positive sign separating the two terms, and six corresponding irrationals with 
a negative sign. These are of course roots of the equations 
x* ± 2a* 3 . p' + (3 . p* - o. 
These irrationals really come before the others in Euclid's order (x. 56 — 
41 for the positive sign and x. 73 — 78 for the negative sign). As we shall 
see in due course, the straight lines actually found by Euclid are 
r . p + JJb . p, the binomial (ij ix Su'o oyoimtiov) 

and the apotome (aVorop,^), 
which are the positive roots of the biquadratic (reducible to a quadratic) 
x>- 2 (1+^p 1 . * s + (r -A)V = °- 

2, i*p ± 6*p, the first bimedial (Ik Svo fito-wv irpcuny) 
and the first apotomc of a medial {p,«7ijs a-voTo/ni) irpuinf), 

which are the positive roots of 

X* - 2 Jk (1 + k) p s . X> + k( I - A)V = o. 

3. k^p + ^r P, the second bimedial (in Buo p-iwav Stvripa) 

and the second apotomc of a medial (/i«n;t (nroTo/17 Sivrtpa), 
which are the positive roots of the equation 

. k + \ . , jk-Xf , 



p 



s/ i *jm t %'/ 1 



Ti * 



Jz V Ji + ^-^zV 7T+-f 

the major (irrational straight line) (ficifav) 
and the m/nw (irrational straight line) ((Wm»), 
which are the positive roots of the equation 

X* - zp a . X 1 + r; p' = o. 

r t + k 1 r 

the "side" of a rational plus a medial (area) (pip-ov m! fUoov Svra/xmj) 
and the " *£& " of a medial minus a rational area (in the Greek 7 /wni p^roS 
jiitrov to oAor irotovcra), 

which are the positive roots of the equation 

2 JP 

'** " ~7=S P* • ** + V.T^ i p * = °' 



fi ^P / k~ A*p /" 



the "side" of the sum of two medial areas {$ Suo /item BumpoTj) 
and the " j<aV " a/ a medial minus a medial area (in the Greek ij pitra jUoou 

pVcTOP to 0A0* TTOlOWa), 

which are the positive roots of the equation 

X l ~2jk.X 1 p'+\ -sp' = 0. 



8 BOOK X 

The above facts and formulae admit of being stated in a great variety of 
ways according to the notation and the particular letters used. Consequently 
the summaries which have been given of Eucl. x. by various writers differ 
much in appearance while expressing the same thing in substance. The first 
summary in algebraical form (and a very elaborate one) seems to have been 
that of Cossali (Origin:, trasporto in Italia, primi progressi in essa del- 
f Algebra, Vol. IL, pp. 242 — 65) who takes credit accordingly (p. 265). In 
1794 Meier Hirsch published at Berlin an Algebraischcr Commcntar iiber das 
sehente Buck der Ekmente des Euklides which gives the contents in algebraical 
form but fails to give any indication of Euclid's methods, using modern forms 
of proof only. In 1834 Poselger wrote a paper, Ueber das zthnte Buck der 
Eletnente des Euklides, in which he pointed out the defects of Hirsch's repro- 
duction and gave a summary of his own, which however, though nearer to 
Euclid's form, is difficult to follow in consequence of an elaborate system of 
abbreviations, and is open to the objection that it is not algebraical enough 
to enable the character of Euclid's irrationals to be seen at a glance. Other 
summaries will be found (1) in Nesselmann, Die Algebra der Griecken, 
pp. 165 — 84; (2) in Loria, Le scienze esatte nelF antica Grecia, 19 14, 
pp. 221—34 > (3) m Christensen's article "Ueber Gleichungen vierten Grades 
im zehnten Buch der Elemente Euklids " in the Zeitsehri/t fiir Math, u, 
Bhysik (Historisek-litteraristhe Abtneitung), xxxtv. (1889), pp. 201 — 17. The 
only summary in English that I know is that in the Benny Cyclopaedia, under 
" Irrational quantity," by De Morgan, who yielded to none in his admiration of 
Book x. " Euclid investigates," says De Morgan, " every possible variety of 
lines which can be represented by J(Ja ± ^/b), a and b representing two 
commensurable lines. ...This book has a completeness which none of the 
others {not even the fifth) can boast of : and we could almost suspect that 
Euclid, having arranged his materials in his own mind, and having completely 
elaborated the loth Book, wrote the preceding books after it and did not live 
to revise them thoroughly." 

Much attention was given to Book x. by the early algebraists. Thus 
Leonardo of Pisa (fl. about 1 200 a.d.) wrote in the 14th section of his Liber 
Abaci on the theory of irrationalities (de tractatu binomiorum et recisorum), 
without however {except in treating of irrational trinomials and cubic irra- 
tionalities) adding much to the substance of Book x. ; and, in investigating 
the equation 

3? + 2X* + I OX - 20, 

propounded by Johannes of Palermo, he proved that none of the irrationals 
in Eucl. x. would satisfy it (Hankel, pp. 344 — 6, Cantor, n„ p. 43). Luca 
Paciuolo (about 1443 — 1514A.D.) in his algebra based himself largely, as he 
himself expressly says, on Euclid x. (Cantor, n,, p. 293). Michael Stifel 
(i486 or 1487 to 1567) wrote on irrational numbers in the second Book of 
his Arithmetica Integra, which Book may be regarded, says Cantor {][,, p. 401), 
as an elucidation of Eucl. x. The works of Cardano (1501 — 76) abound in 
speculations regarding the irrationals of Euclid, as may be seen by reference to 
Cossali (Vol. 11., especially pp. 268—78 and 383—99); the character of 
the various odd and even powers of the binomials and apotomes is therein 
investigated, and Cardano considers in detail of what particular forms of equa- 
tions, quadratic, cubic, and biquadratic, each class of Euclidean irrationals can 
be roots. Simon Stevin (^48—1620) gave an Appendice des incommensurables 
grandeurs en laqudle est sommairement (Uclari le content* du Dixiesme Livre 
d'Euclide (Oevirrcs mathe"matiques, Leyde, 1634, pp. 218-22); he speaks thus 



INTRODUCTORY NOTE 9 

of the book : " La difficulty du dixiesme Livre d'Euclide est a plusieurs 
devenue en horreur, voire jusque & I'appeler la croix des mathematiciens, 
matiere trop dure a digger, et en la quelle n'apercpivent aucune utility," a 
passage quoted by Ijoria (op. at, p. 222). 

It will naturally be asked, what use did the Greek geometers actually 
make of the theory of irrationals developed at such length in Book x. ? The 
answer is that Euclid himself, in Book X111., makes considerable use of the 
second portion of Book x. dealing with the irrationals affected with a negative 
sign, the apotomes etc. One object of Book xin. is to investigate the relation 
of the sides of a pentagon inscribed in a circle arid of an icosahedron and 
dodecahedron inscribed in a sphere to the diameter of the circle or sphere 
respectively, supposed rational. The connexion with the regular pentagon of 
a straight line cut in extreme and mean ratio is well known, and Euclid first 
proves (xm. 6} that, if a rational straight line is so divided, the parts are the 
irrationals called apotomes, the lesser part being a first apotome. Then, on 
the assumption that the diameters of a circle and sphere respectively are 
rational, he proves (xin. n) that the side of the inscribed regular pentagon is 
the irrational straight line called minor, as is also the side of the inscribed 
icosahedron (xin. 16), while the side of the inscribed dodecahedron is the 
irrational called an apoiomt (xin. 17}, 

Of course the investigation in Book x, would not have been complete if 
it had dealt only with the irrationals affected with a negative sign. Those 
affected with the positive sign, the binomials etc., had also to be discussed, 
and we find both portions of Book x., with its nomenclature, made use of by 
Pappus in two propositions, of which it may be of interest to give th*- enun- 
ciations here. 

If, says Pappus (iv. p. 178), A3 be the rational diameter of a semicircle, and 
if AB be produced to C so that B C is equal to the radius, if CD be a tangent, 




if E be the middle point of the arc BD, and if CE be joined, then CE is the 
irrational straight line called minor. As a matter of fact, if p is the radius, 

If, again {p. 182), CD be equal to the radius of a semicircle supposed 




rational, and if the tangent DB be drawn and the angle ADB be bisected by 
DP meeting the circumference in F, then DF is the excess by which the 
binomial exceeds the straight line which produces with a rational area a medial 



IO BOOK X [x. DEKF, 1—4 

whole (see EucL x. 77), {In the figure DK is the binomial and .O" the other 
irrational straight line.) As a matter of fact, if p be the radius, 

Proclus tells us that Euclid left out, as alien to a selection of elements, the 
discussion of the more complicated irrationals, "the unordered irrationals which 
Apollonius worked out more fully" (Proclus, p. 74, 23), while the scholiast 
to Book X. remarks that Euclid does not deal with all rational s and irrationals 
but only the simplest kinds by the combination of which an infinite number 
of irrationals are obtained, of which Apollonius also gave some. The author 
of the commentary on Book x. found by Woepcke in an Arabic translation, 
and above alluded to, also says that "it was Apollonius who, beside the 
ordered irrational magnitudes, showed the existence of the unordered ^nd by 
accurate methods set forth a great number of them." It can only be vaguely 
gathered, from such hints as the commentator proceeds to give, what the 
character of the extension of the subject given by Apollonius may have been. 
See note at end of Book. 



DEFINITIONS. 

i. Those magnitudes are said to be commensurable 
which are measured by the same measure, and those incom- 
mensurable which cannot have any common measure. 

2. Straight lines are commensurable in square when 
the squares on them are measured by the same area, and 
incommensurable in square when the squares on them 
cannot possibly have any area as a common measure. 

3. With these hypotheses, it is proved that there exist 
straight lines infinite in multitude which are commensurable 
and incommensurable respectively, some in length only, and 
others in square also, with an assigned straight line. Let 
then the assigned straight line be called rational, and those 
straight lines which are commensurable with it, whether in 
length and in square or in square only, rational, but those 
which are incommensurable with it irrational, 

4. And let the square on the assigned straight line be 
called rational and those areas which are commensurable 
with it rational, but those which are incommensurable with 
it irrational, and the straight lines which produce them 
irrational, that is, in case the areas are squares, the sides 
themselves, but in case they are any other rectilineal figures, 
the straight lines on which are described squares equal to 
them. 



x. dkff. i— 3 ] DEFINITIONS AND NOTES n 



Definition i. 

%i'p.ificTpa pfyiOrj Xiyttat. ra ttZ avry t** T p*p ptftptrvfitva, AtrvfLptTpa &€, w 
^^div /vSi^«t<u kqivov ftirpov ytvirr&at 



Definition 2. 

KvO< tat owdfLU 0-vftji.tTpoi tUrtVf otav to. air* avrwv T<rpayu*a rtp a&rtp \wptw 
fi.trpt]ttxt t antJ^/urpot 8% OTav tow air' avrw T«Tpayti>fot9 ftYfStv cvSifflTal ^wpioj' 
notfof nit pin' yiriuBtu. 

Commensurable in square is in the Greek Wa/u« o-i^trpM. In earlier 
translations (e.g. Williamson's) 6i>ia/i« has been translated " in power," but, 
as the particular power represented by Swttpci in Greek geometry is square, 
I have thought it best to use the latter word throughout. It will be observed 
that Euclid's expression commensurable in square only (used in Def. 3 and 
constantly) corresponds to what Plato makes Theaetetus call a square root 
(Silt-nun) in the sense of a surd. If a is any straight line, a and ajm, or 
a^Jm and a>jn (where m, n are integers or arithmetical fractions in their 
lowest terms, proper or improper, but not square) are commensurable in square 
only. Of course (as explained in the Porism to X. 10) all straight lines 
commensurable in length ((irjKtt), in Euclid's phrase, are commensurable in 
square also ; but not all straight lines which are commensurable in square are 
commensurable in lengtk as well. On the other hand, straight lines incom- 
mensurable in square are necessarily incommensurable in length also ; but not 
all straight lines which are incommensurable in length are incommensurable 
in square. In fact, straight lines which are commensurable in square only are 
incommensurable in length, but obviously not incommensurable in square. 



Definition 3. 

TfWTGJV faOKUIlilriOV hf.tKVVta.lf OTl Tft irport&tfan] tvBf.it;. vTt^pyovtrw tvOtiai 
TvkrjOtL axttpot trvfiftttpot tt Ktu aui'fifjutptn at fiiv p-rjKtt ftovov, at S* Kal (Wti/te t. 
xaAiiirStii ovr ij ftcy rptrri$tura li&im fn}rq, Kal til ruiJrj; froji/ittpoi tin /XTjKtt xal 
owdjjLtt t iTt Stjraacr futvoy fiip-fit, at di tavrvj axrvftfittTpoi tlAoyut ku A<ttr#tifirav. 

The first sentence of the definition is decidedly elliptical. It should, 
strictly speaking, assert that " with a given straight line there are an infinite 
number of straight lines which are (t) commensurable either (a) in square 
only or (b) in square and in length also, and (2) incommensurable, either 
(a) in length only or (b) in length and in square also." 

The relativity of the terms rational and irrational is well brought out in 
this definition. We may set out any straight line and call it rational, and it 
is then with reference to this assumed rational straight line that others are 
called rational or irrational. 

We should carefully note that the signification of rationale Euclid is wider 
than in our terminology. With him, not only is a straight tine commensurable in 
length with a rational straight line rational, but a straight line is rational which 
is commensurable with a rational straight line in square only. That is, if p is a 

rational straight line, not only is — p rational, where m, n are integers and 



ia BOOK X [x. deff. 3, 4 

mjn in its lowest terms is not square, but */ — .pis rational also. We should 

in this case call . / — . p irrational. It would appear that Euclid's termino- 
logy here differed as much from that of his predecessors as it does from 
ours. We are familiar with the phrase apfajros hiAfurptn rift irt/urciSM by 
which Plato {evidently after the Pythagoreans) describes the diagonal of a 
square on a straight line containing 5 units of length. This " inexpressible 
diameter of five (squared) " means V50, in contrast to the ptp-7 &ap.cTp<«, the 
" expressible diameter " of the same square, by which is meant the approxi- 
mation ^50-1, or 7. Thus for Euclid's predecessors \/^.p would 

apparently not have been rational but apptyrm, " inexpressible," i.e. irrational. 

I shall throughout my notes on this Book denote a rational straight line in 
Euclid's sense by p, and by p and cr when two different rational straight lines are 
required. Wherever then I use p or tr, it must be remembered that p, cr may 
have either of the forms a, ^'i . a, where a represents a units of length, a being 
either an integer or of the form m' n, where m, n are both integers, and k is an 
integer or of the form mjn (where both m, n are integers) but not square. In 
other words, p, ir may have either of the forms a or J A, where A represents 
A units of arm and A is integral or of the form mjn, where m, n are both 
integers. It has been the habit of writers to give a and J a as the alternative 
forms of p, but I shall always use J A for the second in order to keep the 
dimensions right, because it must be borne in mind throughout that p is an 
irrational straight linr. 

As Euclid extends the signification of rational (/Wtos, literally expressible), 
so he limits the scope of the term dAo-yov (literally having no ratio) as applied 
to straight lines. That this limitation was started by himself may perhaps be 
inferred from the form of words " let straight lines incommensurable with it 
be tailed irrational." Irrational straight lines then are with Euclid straight lines 
commensurable neither in length nor in square with the assumed rational 
straight line- Jk . a where k is not square is not irrational; $k . a is irrational, 
and so (as we shall see later on) is (y/6± s/*)a. 



Definition 4. 

Kat to pep airo T17S TrpaTiQitoifi tv&tiiis TCTpayowov pyrov, «ai t« rounp 
(rvfijarpa /n^Tit, ra Si Toi-rw atrvp/icrpa dAoya Kakiitrdv, kcu ai cWa/Kwu aura 
aAoyot, (i piv T«Tpay*>™ *"), aurai at irAtvpai, *£ St trtpa riva ivffiiypap-pa,, ai 
ttra avrot? TCTpayaiva apaypa<^owrai. 

As applied to areas, the terms rational and irrational have, on the other 
hand, the same sense with Euclid as we should attach to them. According 
to Euclid, if p is a rational straight line in his sense, p 1 is rational and any 
area commensurable with it, i.e. of the form kp* (where k is an integer, or of 
the form mjn, where m, n are integers), is rational ; but any area of the form 
n/h . p* is irrational. Euclid's rational area thus contains A units of area, 
where A is an integer or of the form mjn, where m, n are integers ; and his 
irrational area is of the form ^/h.A. His irrational area is then connected 
with his irrational straight line by making the latter the square root of the 



DEF. 4] 



NOTES ON DEFINITIONS 3, 4 



•3 



former. This would give us for the irrational straight line if A . J A, which of 
course includes ijk.a. 

at BwafitKu aura are the straight lines the squares on which are equal to 
the areas, in accordance' with the regular meaning of SvyatrOai, It is scarcely 
possible, in a book written in geometrical language, to translate Suwi/ifn) as 
the square root (of an area) and Jwmrfai as to be the square root (of an area), 
although I can use the term " square root " when in my notes I am using an 
algebraical expression to represent an area ; I shall therefore hereafter use the 
word "side" for Swa/ttVij and "10 be the side of" for huvaaOai, so that 
"side" will in such expressions be a short way of expressing the "side of 
a square equal to (an area)." In this particular passage it is not quite practi- 
cable to use the words " side of" or " straight line the square on which is equal 
to," for these expressions occur just afterwards for two alternatives which the 
word SumijmVi) covers. I have therefore exceptionally translated " the straight 
lines which produce them " (i.e. if squares are described upon them as sides). 

<ii "uro. avruw TtTjiiytava. avaypu^Hiiwat, literally " the (straight lines) which 
describe squares equal to them" : a peculiar use of the active of draypdijmr, 
the meaning being of course " the straight lines on which are described the 
squares" which are equal to the rectilineal figures. 















BOOK X. PROPOSITIONS. 



Proposition 



Two unequal magnitudes being set out, if from the greater 
there be subtracted a magnitude greater than its half, and from 
that -which is left a magnitude greater than its half and if 
this process be repeated continually, there will be left some 
magnitude which will be less than the lesser magnitude set out. 

Let AB, C be two unequal magnitudes of which AB is 
the greater : 

I say that, if from AB there be * — ' ' B ° 

subtracted a magnitude greater o -t + e 

than its half, and from that which 

is left a magnitude greater than its half, and if this process be 
repeated continually, there will be left some magnitude which 
will be less than the magnitude C. 

For C if multiplied will sometime be greater than AB. 

[cf. v, Def. 4] 

Let it be multiplied, and let DE be a multiple of C, and 
greater than. AB ; 

let DE be divided into the parts DE, EG, GE equal to C, 
from AB let there be subtracted BH greater than its half, 
and, from AH, HK greater than its half, 
and let this process be repeated continually until the divisions 
in AB are equal in multitude with the divisions in DE. 

Let, then, AK, KH, HB be divisions which are equal in 
multitude with DE, EG, GE. 

Now, since DE is greater than AB, 
and from DE there has been subtracted EG less than its 
half, 

and, from AB, BH greater than its half, 
therefore the remainder GD is greater than the remainder HA. 



x. i] PROPOSITION I IS 

And, since GD is greater than HA, 
and there has been subtracted, from GD, the half GP, 
and, from HA, HK greater than its half, 
therefore the remainder DF\s greater than the remainder A K. 

But DF is equal to C ; 
therefore C is also greater than A K. 

Therefore AK is less than C. 

Therefore there is left of the magnitude AB the magnitude 
AK which is less than the lesser magnitude set out, namely C. 

Q. E. D. 

And the theorem can be similarly proved even if the parts 
subtracted be halves. 

This proposition will be remembered because it is the lemma required in 
Euclid's proof of XII. 2 to the effect that circles are to one another as the 
squares on their diameters. Some writers appear to be under the impression 
that xii. 2 and the other propositions in Book xii. in which the method of 
exhaustion is used are the only places where Euclid makes use of X. t ; and it 
is commonly remarked that x. 1 might just as well have been deferred till the 
beginning of Book XII. Even Cantor (Geseh. d. Math, ij, p. 269) remarks 
that " Euclid draws no inference from it [x. 1], not even that which we should 
more than anything else expect, namely that, if two magnitudes are incom- 
mensurable, we can always form a magnitude commensurable with the first 
which shall differ from the second magnitude by as little as we please." But, 
so far from making no use of x. 1 before xii. 2, Euclid actually uses it in the 
very next proposition, x. 2. This being so, as the next note will show, it 
follows that, since x. 2 gives the criterion for the incommensurability of two 
magnitudes (a very necessary preliminary to the study of in com m ens u rabies), 
X: 1 comes exactly where it should be. 

Euclid uses X. 1 to prove not only xn. 2 but xii. 5 (that pyramids with the 
same height and triangular bases are to one another as their bases), by means 
of which he proves (xn. 7 and Por.) that any pyramid is a third part of the 
prism which has the same base and equal height, and xn. 10 (that any cone 
is a third part of the cylinder which has the same base and equal height), 
besides other similar propositions. Now xn. 7 Por. and xii. 10 are theorems 
specifically attributed to Eudoxus by Archimedes (On the Sphere and Cylinder, 
Preface), who says in another place (Quadrature of the Parabola, Preface) that 
the first of the two, and the theorem that circles are to one another as the 
squares on their diameters, were proved by means of a certain lemma which 
he states as follows : "Of unequal lines, unequal surfaces, or unequal solids, 
the greater exceeds the less by such a magnitude as is capable, if added 
[continually] to itself, of exceeding any magnitude of those whfch are 
comparable with one another," i.e. of magnitudes of the same kind as the 
original magnitudes. Archimedes also says (he. eit.) that the second of 
the two theorems which he attributes to Eudoxus (End. xii. 10) was 
proved by means of "a lemma similar to the aforesaid." The lemma 
stated thus by Archimedes is decidedly different from x. i, which, however, 
Archimedes himself uses several times, while he refers to the use of it 



16 BOOK X [x. i 

in xii. 2 {On the Sphere and Cylinder, I. 6). As I have before suggested 
{ The Works of Archimedes, p. xlviii), the apparent difficulty caused by the 
mention of two lemmas in connexion with the theorem of Eucl. xn. 2 may be 
explained by reference to the proof of x. 1. Euclid there takes the lesser 
magnitude and says that it is possible, by multiplying it, to make it some time 
exceed the greater, and this statement he clearly bases on the 4th definition of 
Book v., to the effect that "magnitudes are said to bear a ratio to one another 
which can, if multiplied, exceed one another." Since then the smaller 
magnitude in x. 1 may be regarded as the difference between some two 
unequal magnitudes, it is clear that the lemma stated by Archimedes is in 
substance used to prove the lemma in x. 1, which appears to play so much 
larger a part in the investigations of quadrature and cubature which have come 
down to us. 

Besides being employed in Eucl. x. 1, the "Axiom of Archimedes" appears 
in Aristotle, who also practically quotes the result of x. 1 itself. Thus he 
says, Physics vm. 10, 266 b 2, " By continually adding to a finite (magnitude) 
1 shall exceed any definite (magnitude), and similarly by 'continually subtract- 
ing from it I shall arrive at something less than it," and Hid. til. 7, 207 b 10 
" For bisections of a magnitude are endless." It is thus somewhat misleading 
to use the term "Archimedes' Axiom" for the "lemma" quoted by him, 
since he makes no claim to be the discoverer of it, and it was obviously much 
earlier. 

Stolz (see G. Vitali in Questioni riguardanti le matematiche elemenfari, 1., 
pp. 1 29—30) showed how to prove the so-called Axiom or Postulate of Archi- 
medes by means of the Postulate of Dedekind, thus. Suppose the two magni- 
tudes to be straight lines. It is required to prove that, given two straight lines, 
(here always exists a multiple 0/ the smaller which is greater than the other. 

Let the straight lines be so placed that they have a common extremity and 
the smaller lies along the other on the same side of the common extremity. 

If AC be the greater and AB the smaller, we have to prove that there 
exists an integral number n such that n. AB> AC. 

Suppose that this is not true but that there are some points, like B, not 
coincident with the extremity A, and such that, n being any integer however 
great, n . AB-cAC; and we have to prove that this assumption leads to an 
absurdity. 

h iyi k i 



The points of AC may be regarded as distributed into two "parts," namely 

(1) points //for which there exists no integer n such that «'. AH> AC-, 

(2) points K for which an integer n does exist such that n, AK> AC. 

This division into parts satisfies the conditions for the application of 
Dedekind's Postulate, and therefore there exists a point M such that the 
points of AM belong to the first part and those of MC to the second part. 

Take now a point Kon MC such that MY< AM. The middle point (X) 
of A Kwill fall between A and M and will therefore belong to the first part; 
but, since there exists an integer a such that n . AV> AC, it follows that 
2n . AX>- AC: which is contrary to the hypothesis. 



x. 2] PROPOSITIONS i, i 17 

Proposition 2. 

If when the less of two unequal magnitudes is continually 
subtracted in turn from the greater, that which is left never 
measures the one before it, the magnitudes will be incom- 
mensurable. 

For, there being two unequal magnitudes AB, CD, and 
AB being the less, when the less is continually subtracted 
in turn from the greater, let that which is left over never 
measure the one before it ; 
I say that the magnitudes AB, CD are incommensurable. 



c £ o 

For, if they are commensurable, some magnitude will 
measure them. 

Let a magnitude measure them, if possible, and let it be E ; 
let AB, measuring FD, leave CT^less than itself, 
let CF measuring BG, leave AG less than itself, 
and let this process be repeated continually, until there is left 
some magnitude which is less than E. 

Suppose this done, and let there be left AG less than E. 

Then, since E measures AB, 
while AB measures DF, 
therefore E will also measure FD. 

But it measures the whole CD also ; 
therefore it will also measure the remainder CF. 

But CF measures BG ; 
therefore E also measures BG. 

But it measures the whole AB also ; 
therefore it will also measure the remainder AG, the greater 
the less : 
which is impossible. 

Therefore no magnitude will measure the magnitudes AB, 
CD; 
therefore the magnitudes AB, CD are incommensurable. 

[x. Def. 1] 

Therefore etc. 



i8 BOOK. X [x. * 

This proposition states the test for incommensurable magnitudes, founded 
on the usual operation for rinding the greatest common measure. The sign 
of the incommensurability of two magnitudes is that this operation never 
comes to an end, while the successive remainders become smaller and smaller 
until they are less than any assigned magnitude. 

Observe that Euclid says " let this process be repeated continually until 
there is left some magnitude which is less than E." Here he evidently 
assumes that the process will some time produce a remainder less than any 
assigned magnitude E. Now this is by no means self-evident, and yet 
Heiberg (though so careful to supply references) and Lorenz do not refer to 
the basis of the assumption, which is in reality x. I, as Billingsley and 
Williamson were shrewd enough to see. The fact is that, if we set off a 
smaller magnitude once or oftcner along a greater which it does not exactly 
measure, until the remainder is less than the smaller magnitude, we take away 
from the greater more than its half. Thus, in the figure, FD is more than the 
half of CD, and BG more than the half of AB. If we continued the process, 
AG marked off along CF as many times as possible would cut off more than 
its half; next, more than half AG would be cut off, and so on. Hence along 
CD, AB alternately the process would cut off more than half, then more than 
half the remainder and so on, so that on both lines we should ultimately 
arrive at a remainder less than any assigned length. 

The method of finding the greatest common measure exhibited in this 
proposition and the next is of course again the same as that which we use and 
which may be shown thus : 

b)u(p 

Pi 
')*(? 

d)c(r 
rd 



The proof too is the same as ours, taking just the same form, as shown in the 
notes to the similar propositions vii. i, 2 above. In the present case the 
hypothesis is that the process never stops, and it is required to prove that a, b 
cannot in that case have any common measure, as/ For suppose that/ is a 
common measure, and suppose the process to be continued until the remainder 
t, say, is less than/ 

Then, since /measures a, b, it measures a -pb, or c. 

Since/ measures b, c, it measures b-qc, or d; and, since/ measures c, d, 
it measures c—rd, or e \ which is impossible, since e<f. 

Euclid assumes as axiomatic that, if/ measures a, />, it measures ma ± nb. 

In practice, of course, it is often unnecessary to carry the process far in 
order to see that it will never stop, and consequently that the magnitudes are 
incommensurable. A good instance is pointed out by Allman (Greek Geometry 
from Thates to Euclid, pp. 4a, 137—8). Euclid proves in XHI. 5 that, if AB 
be cut in extreme and mean ratio at C, and if 

DA equal to AC be added, then DB is also cut D A c B 

in extreme and mean ratio at A. This is indeed 

obvious from the proof of 11. t r. It follows conversely that, if BD is cut into 
extreme and mean ratio at A, and AC, equal to the lesser segment AD, he 
subtracted from the greater AB, AB is similarly divided at C. We can then 



x. s] PROPOSITION a 19 

mark off from AC a. portion equal to CB, and AC will then be similarly 
divided, and so on. Now the greater segment in a line thus divided is greater 
than half the line, but it follows from xni. 3 that it is less than twice the 
lesser segment, i.e. the lesser segment can never be marked off more than 
once from the greater. Our process of marking off the lesser segment from the 
greater continually is thus exactly that of finding the greatest common measure. 
If, therefore, the segments were commensurable, the process would stop. But 
it clearly does not ; therefore the segments are incommensurable. 

Allman expresses the opinion that it was rather in connexion with the line 
cut in extreme and mean ratio than with reference to the diagonal and side 
of a square that the Pythagoreans discovered the incommensurable. But the 
evidence seems to put it beyond doubt that the Pythagoreans did discover 
the incommensurability of J2 and devoted much attention to this particular 
case. The view of Allman does not therefore commend itself to me, though 
it is likely enough that the Pythagoreans were aware of the incommensura- 
bility of the segments of a line cut in extreme and mean ratio. At all events 
the Pythagoreans could hardly have carried their investigations into the in- 
commensurability of the segments of this line very far, since Theaetetus is 
said to have made the first classification of irrationals, and to him is also, with 
reasonable probability, attributed the substance of the first part of Eucl. xni., 
in the sixth proposition of which occurs the proof that the segments of a 
rational straight line cut in extreme and mean ratio are apotomes. 

Again, the incommensurability of Ji can be proved by a method 
practically equivalent to that of x. 2, and without carrying the process very 
far. This method is given in Chrystal's Text- 
book of Algebra (1. p. 270). Let d, a be the 
diagonal and side respectively of a square 
A BCD. Mark off AF along A C equal to a. 
Draw FE at right angles to AC meeting BC 
in E. 

It is easily proved that 

BE = EF= FC, ©/ 

CF=AC-AB = d-a (r). v 

CE=C3- CF=a-{d-a) 

= xa-d (z). 

Suppose, if possible, that d, a are commensurable. If d, a are both 
commensurably expressible in terms of any finite unit, each must be an 
integral multiple of a certain finite unit 

But from (1) it follows that CF, and from (2) it follows that CE, is an 
integral multiple of the same unit. 

And CF, CE are the side and diagonal of a square CFEG, the side of 
which is less than half the side of the original square. 1 f a u d x are the side and 
diagonal of this square, 

di = ia-d I ' 

Similarly we can form a square with side a % and diagonal d, which are less 
than half a„ d, respectively, and a„ d, must be integral multiples of the same 
unit, where 

a t = d x -a lt 




so BOOK X [» 2, 3 

and this process may be continued indefinitely until (x. i) we have a square 
as small as we please, the side and diagonal of which are integral multiples of 
a finite unit : which is absurd. 

Therefore a, d are incommensurable. 

It will be observed that this method is the opposite of that shown in the 
Pythagorean series of side- and diagonal-numbers, the squares being 
successively smaller instead of larger. 



Proposition 3. 

Given two commensurable magnitudes, to find their greatest 
common measure. 

Let the two given commensurable magnitudes be AB, CD 
of which AB is the less ; 

thus it is required to find the greatest common measure of 
AB, CD. 

Now the magnitude AB either measures CD or it does 
not. 

If then it measures it — and it measures itself also — AB is 
a common measure of AB, CD, 

And it is manifest that it is also the greatest ; 
for a greater magnitude than the magnitude AB will not 
measure AB. 

-a 
-a 



-9- .4 



Next, let AB not measure CD. 

Then, if the less be continually subtracted in turn from 
the greater, that which is left over will sometime measure 
the one before it, because AB, CD are not incommensurable; 

[cf. x. 2] 
let AB, measuring ED, leave EC less than itself, 

let EC, measuring FB, leave AF less than itself, 
and let AF measure CE. 

Since, then, AF measures CE, 
while CE measures FB, 
therefore AF will also measure FB. 

But it measures itself also ; 
therefore AF will also measure the whole AB. 



x. 3] PROPOSITIONS a, 3 ai 

But AB measures DE ; 
therefore AF will also measure ED. 

But it measures CE also ; 
therefore it also measures the whole CD. 

Therefore AF is a common measure of AB, CD. 

I say next that it is also the greatest. 

For, if not, there will be some magnitude greater than AF 
which will measure AB, CD. 

Let it be G. 

Since then G measures AB, 
while AB measures ED, 
therefore G will also measure ED. 

But it measures the whole CD also ; 
therefore G will also measure the remainder CE. 

But CE measures FB ; 
therefore G will also measure FB. 

But it measures the whole AB also, 
and it will therefore measure the remainder AF, the greater 
the less : 
which is impossible. 

Therefore no magnitude greater than AF will measure 
AB, CD; 
therefore AF'is the greatest common measure of AB, CD. 

Therefore the greatest common measure of the two given 
commensurable magnitudes AB, CD has been found. 

Q. E. D. 

Porism. From this it is manifest that, if a magnitude 
measure two magnitudes, it will also measure their greatest 
common measure. 

This proposition for two commensurable magnitudes is, mutatis mutandis, 
exactly the same as vu. 2 for numbers. We have the process 

b)a{p 
pb_ 

T)b(q 
qc 

rd 
where c is equal to rd and therefore there is no remainder, 



« BOOK X [x. 3, 4 

It is then proved that d is a common measure of a, 6; and next, by a 
reiurtio ad adsardum, that it is the greatat common measure, since any 
common measure must measure d, and no magnitude greater than d can 
measure d. The rtductw ad abmrdum is of course one of form only. 

The Porism corresponds exactly to the Porism to vn. a. 

The process of finding the greatest common measure is probably given in 
this Book, not only for the sake of completeness, but because in x. 5 a 
common measure of two magnitudes A, B is assumed and used, and therefore 
it is important to show that such a measure can be found if not already 
known. 

Proposition 4. 

Given three commensurable magnitudes, to find their greatest 
common measure. 

Let A, B, C be the three given commensurable magnitudes; 
thus it is required to find the greatest 
common measure of A, B, C. A 

Let the greatest common measure b 

of the two magnitudes A, B be taken, c- 

and let it be D ; [x. 3] D E F 

then D either measures C, or does 
not measure it. 

First, let it measure it. 

Since then D measures C, 
while it also measures A, B, 
therefore D is a common measure of A, B, C. 

And it is manifest that it is also the greatest ; 
for a greater magnitude than the magnitude D does not 
measure A, B. 

■ 

Next, let D not measure C. 

\ say first that C, D are commensurable. 

For, since A, B, C are commensurable, 

some magnitude will measure them, 

and this will of course measure A, B also ; 

so that it will also measure the greatest common measure of 
A, B, namely D. [x. 3, Por.] 

But it also measures C; 

so that the said magnitude will measure C, D ; 

therefore C, D are commensurable. 



x. 4] PROPOSITIONS 3l 4 23 

Now let their greatest common measure be taken, and let 
it be E. [x. 3] 

Since then E measures D, 
while D measures A, B t 
therefore E will also measure A, B. 

But it measures C also ; 
therefore E measures A, B, C ; 
therefore E is a common measure of A, B, C. 

I say next that it is also the greatest. 

For, if possible, Jet there be some magnitude F greater than 
E, and let it measure A, B, C. 

Now, since F measures A, B, C, 
it will also measure A, B, 
and will measure the greatest common measure of A, B. 

[x. 3. Por.] 

But the greatest common measure of A, B is D; 

therefore F measures D. 

But it measures C also ; 
therefore F measures C, D ; 

therefore F will also measure the greatest common measure 
of C, D. [x, 3, Por.] 

But that is E; 
therefore F will measure E, the greater the less : 
which is impossible. 

Therefore no magnitude greater than the magnitude E 
will measure A, B, C; 

therefore E is the greatest common measure of A, B, C \i D 
do not measure C, 
and, if it measure it, D is itself the greatest common measure. 

Therefore the greatest common measure of the three given 
commensurable magnitudes has been found. 

Porism. From this it is manifest that, if a magnitude 
measure three magnitudes, it will also measure their greatest 
common measure. 

Similarly too, with more magnitudes, the greatest common 
measure can be found, and the porism can be extended. 

Q. E. D. 



M 



BOOK X [x. 4, 5 



This proposition again corresponds exactly to vn, 3 for numbers. As 
there Euclid thinks it necessary to prove that, a, b, c not being prime to one 
another, d and c are also not prime to one another, so here he thinks it 
necessary to prove that d, c are commensurable, as they must be since any 
common measure of a, b must be a measure of their greatest common 
measure d (x. 3, Por.). 

The argument in the proof that t, the greatest common measure of d, c, is 
the greatest common measure of a, b, c, is the same as that in vn. 3 and x. 3. 

The Porism contains the extension of the process to the case of four 
or more magnitudes, corresponding to Heron's remark with regard to the 
similar extension of vn. 3 to the case of four or more numbers. 



Proposition 5. 

Commensurable magnitudes have to one another the ratio 
which a number has to a number. 

Let A, B be commensurable magnitudes ; 

I say that A has to B the ratio which a number has to a 
number. 

For, since A, B are commensurable, some magnitude will 
measure them. 

Let it measure them, and let it be C, 



And, as many times as C measures A, so many units let 
there be in D ; 

and, as many times as C measures B, so many units let there 
be in E. 

Since then C measures A according to the units in D, 

while the unit also measures D according to the units in it, 

therefore the unit measures the number D the same number 
of times as the magnitude C measures A ; 

therefore, as C is to A, so is the unit to D ; [vn, Def. *o) 

therefore, inversely, as A is to C, so is D to the unit. 

[cf. v. 7, Por.] 
Again, since C measures B according to the units in E, 
while the unit also measures E according to the units in it, 



x. S ] PROPOSITIONS 4, 5 35 

therefore the unit measures E the same number of times as C 
measures B ; 

therefore, as C is to B, so is the unit to E. 
But it was also proved that, 

as A is to C, so is D to the unit ; 
therefore, ex aequali, 

as A is to B, so is the number D to E. [v. 22J 

Therefore the commensurable magnitudes A, 3 have to 
one another the ratio which the number D has to the number E. 

Q. E. D, 

The argument is as follows. If a, b be commensurable magnitudes, they 
have some common measure c, and 

a-mc t 
b=t«e, 
where m, n are integers. 

It follows that c:a= 1 :m (i), 

or, inversely, a : c = m : t j 

and also that c : b = 1 : «, 

so that, ex aequali, a :b = m;n. 

It will be observed that, in stating the proportion (1), Euclid is merely 
expressing the fact that a is the same multiple of c that m is of 1. In other 
words, he rests the statement on the definition of proportion in vil, Def. 20. 
This, however, is applicable only to four numbers, and c, a are not numbers but 
magnitudes. Hence the statement of the proportion is not legitimate unless 
it is proved that it is true in the sense of v. Det. 5 with regard to magnitudes 
in general, the numbers 1, m being magnitudes. Similarly with regard to the 
other proportions in the proposition. 

There is, therefore, a hiatus. Euclid ought to have proved that magnitudes 
which are proportional in the sense of vn. Def. 10 are also proportional in the 
sense of v. Def. 5, or that the proportion of numbers is included in the 
proportion of magnitudes as a particular case. Simson has proved this in his 
Proposition C inserted in Book v. (see Vol. 11. pp. 1 26 — 8). The portion of 
that proposition which is required here is the proof that, 
if a = mb \ 

c = md)' 
then a 1 &W : % M the sense of v. Def. $ . 

Take any equimultiples pa, pe of a, c and any equimultiples qb, qd of b, d. 

Now pa =pmb 1 

pc=pmd) ' 

But, according as pmb > = <qb, pmd > = <qd. 
Therefore, according as pa > = < qb, pm > = < qd. 

And pa, pc are any equimultiples of a, c, and qb, qd any equimultiples 
of b, d. 

Therefore a:b = e\d. [v, Def. S-] 



26 BOOK X [x. 6 

Proposition 6. 

If two magnitudes have to one another the ratio which a 
number has to a number, the magnitudes will be commensurable. 

For let the two magnitudes A, B have to one another the 
ratio which the number D has to the number E \ 
s I say that the magnitudes A, B are commensurable. 
A -> i b c 



For let A be divided into as many equal parts as there 
are units in D, 

and let C be equal to one of them ; 

and let F be made up of as many magnitudes equal to C as 
io there are units in E. 

Since then there are in A as many magnitudes equal to C 
as there are units in D, 

whatever part the unit is of D, the same part is C of A also ; 
therefore, as C is to A, so is the unit to D. [vn. Def. 20] 

15 But the unit measures the number D ; 
therefore Calso measures A. 

And since, as C is to A, so is the unit to D, 
therefore, inversely, as A is to C, so is the number D to the 
unit. [cf. v. 7, Por.] 

to Again, since there are in F as many magnitudes equal 
to C as there are units in E, 

therefore, as C is to F, so is the unit to E. [vu. Def. 20] 

But it was also proved that, 

as A is to C, so is D to the unit ; 
25 therefore, ex aeouali, as A is to F, so is D to E. [v. 22] 

But, as D is to E, so is A to B ; 
therefore also, as A is to B t so is it to ^also. [v. n] 

Therefore A has the same ratio to each of the magnitudes 
B, F; 
30 therefore B is equal to F. [v. 9] 

But C measures F; 
therefore it measures B also. 

Further it measures A also ; 
therefore C measures A, B. 



x. 6] PROPOSITION 6 27 

3S Therefore A is commensurable with B. 
Therefore etc. 

Porism. From this it is manifest that, if there be two 
numbers, as D, E, and a straight line, as A, it is possible to 
make a straight line \F"\ such that the given straight line is to 
40 it as the number D is to the number E, 

And, if a mean proportional be also taken between A, F, 
as B, 

as A is to F, so will the square on A be to the square on B, 

that is, as the first is to the third, so is the figure on the first 

4s to that which is similar and similarly described on the second. 

[vj. 19, Por.] 
But, as A is to F, so is the number D to the number E\ 
therefore it has been contrived that, as the number D is to 
the number E, so also is the figure on the straight line A to 
the figure on the straight line B. Q. e. d. 

15. But the unit measures the number D ; therefore C also measures A. 
These words are redundant, thmigh they are apparently found in all the MSs. 

The same link to connect the proportion of numbers with the proportion 
of magnitudes as was necessary in the last proposition is necessary here. This 
being premised, the argument is as follows. 

Suppose a : b = m:rt, 

where m, n are (integral) numbers. 

Divide a into m parts, each equal to e, say, 
so that a = mc. 

Now take d such that d = nc. 

Therefore we have a\c-m;\, 

and c : d = 1 : «, 

so that, ex aequali, a :d=m : » 

= a : $, by hypothesis. 

Therefore b-d-- ne, 
so that c measures b n times, and a, b are commensurable. 

The Porism is often used in the later propositions. It follows (1) that, if 
a be a given straight line, and m, n any numbers, a straight line x can be 
found such that 

a : x — m : n. 
(a) We can find a straight line^ such that 

a* s y* = m : n. 
For we have only to take y, a mean proportional between a and x, as 
previously found, in which case a, y, x are in continued proportion and 
[v. Def. 9] 

t^'.y = a : x 
= m : «. 



*8 BOOK X [x. 7—9 

Proposition 7. 

Incommensurable magnitudes have not to one another the 
ratio which a number has to a number. 

Let A, B be incommensurable magnitudes ; 
I say that A has not to B the ratio which a number has to a 
number. 

For, if A has to B the ratio which a number has to a 
number, A will be commensurable with B. [x 6] 

But it is not ; 
therefore A has not to B the ratio which a ^ 

number has to a number. 

Therefore etc. 

Proposition 8. 

If two magnitudes have not to one another the ratio which 
a number has to a number, the magnitudes will be incom- 
mensurable. 

For let the two magnitudes A, B not have to one another 
the ratio which a number has to a number ; 

I say that the magnitudes A, B are incom- — - 

mensurable. ~ 

For, if they are commensurable, A will have to B the 
ratio which a number has to a number. [x. 5] 

But it has not ; 
therefore the magnitudes A, B are incommensurable. 

Therefore etc. 

Proposition 9. 

The squares on straight lines commensurable in length have 
to one another the ratio which a square number has to a square 
number; and squares which have to one another the ratio 
which a square number has to a square number will also have 
their sides commensurable in length. But the squares on 
straight lines incommensurable in length have not to one 
another the ratio which a square number has to a square 
number ; and squares which have not to one another the ratio 
which a square number has to a square number will not have 
their sides commensurable in length either. 



x. 9j PROPOSITIONS 7—9 19 

For let A, B be commensurable in length ; 
I say that the square on A 

has to the square on B the ~ 

ratio which a square number ~~ ^ 

has to a square number. 

For, since A is commensurable in length with B, 
therefore A has to B the ratio which a number has to a 
number. [x. 5] 

Let it have to it the ratio which C has to D. 

Since then, as A is to B, so is C to D, 
while the ratio of the square on A to the square on B is 
duplicate of the ratio of A to B, 

for similar figures are in the duplicate ratio of their corre- 
sponding sides; [vi. zo, Por.] 
and the ratio of the square on C to the square on D is duplicate 
of the ratio of C to D, 

for between two square numbers there is one mean proportional 
number, and the square number has to the square number the 
ratio duplicate of that which the side has to the side ; [vm. n] 
therefore also, as the square on A is to the square on B, so 
is the square on C to the square on D. 

Next, as the square on A is to the square on B, so let 
the square on C be to the square on D ; 
I say that A is commensurable in length with B. 

For since, as the square on A is to the square on B, so is 
the square on C to the square on D, 

while the ratio of the square on A to the square on B is 
duplicate of the ratio of A to B, 

and the ratio of the square on C to the square on D is duplicate 
of the ratio of C to D, 
therefore also, as A is to B, so is C to D. 

Therefore A has to B the ratio which the number C has 
to the number D ; 
therefore A is commensurable in length with B. [x. 6] 

Next, let A be incommensurable in length with B ; 
I say that the square on A has not to the square on B the 
ratio which a square number has to a square number. 

For, if the square on A has to the square on B the ratio 



30 BOOK X [x. 9 

which a square number has to a square number, A will be 
commensurable with B. 

But it is not ; 
therefore the square on A has not to the square on B the 
ratio which a square number has to a square number. 

Again, let the square on A not have to the square on B 
the ratio which a square number has to a square number ; 
I say that A is incommensurable in length with B. 

For, if A is commensurable with B, the square on A will 
have to the square on B the ratio which a square number has 
to a square number. 

But it has not ; 
therefore A is not commensurable in length with B. 

Therefore etc. 

Porism. And it is manifest from what has been proved 
that straight lines commensurable in length are always com- 
mensurable in square also, but those commensurable in square 
are not always commensurable in length also. 

[Lemma. It has been proved In the arithmetical books 
that similar plane numbers have to one another the ratio 
which a square number has to a square number, [vm. 26] 

and that, if two numbers have to one another the ratio which 
a square number has to a square number, they are similar 
plane numbers. [Converse of vm. a6] 

And it is manifest from these propositions that numbers 
which are not similar plane numbers, that is, those which 
have not their sides proportional, have not to one another 
the ratio which a square number has to a square number. 

For, if they have, they will be similar plane numbers : 
which is contrary to the hypothesis. 

Therefore numbers which are not similar plane numbers 
have not to one another the ratio which a square number has 
to a square number.] 

A scholium to this proposition (Schol. x. No. 62) says categorically that 
the theorem proved in it was the discovery of T/heaetetus. 

If a, b be straight lines, and 

a : b = fit : «, 
where m, n are numbers, 
then f-.P^nP-.tt; 

and conversely. 



x. 9, roj PROPOSITIONS 9, 10 31 

This inference, which looks so easy when thus symbolically expressed, was 
by no means so easy for Euclid owing to the fact that a, b are straight lines, 
and m, n numbers. He has to pass from a : b to a 1 : b* by means of vi. 20, Por. 
through the duplicate ratio; the square on a is to the square on b in the 
duplicate ratio of the corresponding sides a, b. On the other hand, mi, m 
being numbers, it is vm. \ 1 which has to be used to show that «' : n % is the 
ratio duplicate of »i : «. 

Then, in order to establish his result, Euclid assumes that, if two ratios art 
equal, the ratios which are their duplicates are also equal. This is nowhere 
proved in Euclid, but it is an easy inference from v. 22, as shown in my note 
on vi. 22. 

The converse has to be established in the same careful way, and Euclid 
assumes that ratios the duplicates of which are equal are themselves equal. 
This is much more troublesome to prove than the converse; for proofs I refer 
to the same note on vi. 22. 

The second part of the theorem, deduced by reductio ad absurdum from 
the first, requires no remark. 

In the Greek text there is an addition to the Porism which Heiberg 
brackets as superfluous and not in Euclid's manner. It consists {1} of a sort 
of proof, or rather explanation, of the Porism and (2) of a statement and 
explanation to the effect that straight lines incommensurable in length are 
not necessarily incommensurable in square also, and that straight lines 
incommensurable in square are, on the other hand, always incommensurable 
in length also. 

The Lemma gives expressions for two numbers which have to one another 
the ratio of a square number to a square number. Similar plane numbers 
are of the form pm . pn and q m . qn, or mnp 1 and mnq 1 , the ratio of which is 
of course the ratio of p* to a*. 

The converse theorem that, if two numbers have to one another the ratio 
of a square number to a square number, the numbers are similar plane 
numbers is not, as a matter of fact, proved in the arithmetical Books. It is 
the converse of vui. 26 and is used in ix. 10. Heron gave it (see note on 
vim. 27 above). 

Heiberg however gives strong reason for supposing the Lemma to be an 
interpolation. It has reference to the next proposition, x. 10, and, as we shall 
see, there are so many objections to x, ro that it can hardly be accepted as 
genuine. Moreover there is no reason why, in the Lemma itself, numbers 
which are not similar plane numbers should be brought in as they are. 



[Proposition 10. 

To find two straight lines incommensurable, the one in 
length only, and the other in square also, with an assigned 
straight line, 

Let A be the assigned straight line ; 
thus it is required to find two straight lines incommensurable, 
the one in length only, and the other in square also, with A. 

Let two numbers B, C be set out which have not to one 



3i BOOK X [x. i© 

another the ratio which a square number has to a square 
number, that is, which are not similar plane 

numbers ; A 

and let it be contrived that, o 

as B is to C, so is the square on A to * " 

the square on D ^^^__ 

— for we have learnt how to do this — 

[x. 6, Por.] 

therefore the square on A is commensurable with the square 
on D. [x. 6] 

And, since B has not to C the ratio which a square number 
has to a square number, 

therefore neither has the square on A to the square on D the 
ratio which a square number has to a square number ; 
therefore A is incommensurable in length with D. [x. 9] 

Let E be taken a mean proportional between A, D ; 

therefore, as A is to D, so is the square on A to the square 
on E. [v. Def. 9] 

But A is incommensurable in length with D ; 

therefore the square on A is also incommensurable with the 
square on E ; [x. 11] 

therefore A is incommensurable in square with E. 

Therefore two straight lines D, E have been found in- 
commensurable, D in length only, and E in square and of 
course in length also, with the assigned straight line A,~\ 

It would appear as though this proposition was intended to supply a 
justification for the statement in x. Def. 3 that it is proved that there are an 
infinite number of straight lines (a) incommensurable in length only, or 
commensurable in square only, and (A) incommensurable in square, with any 
given straight line. 

But in truth the proposition could well be dispensed with ; and the 
positive objections to its genuineness are considerable. 

In the first place, it depends on the following proposition," x. 1 1 ; for the 
last step concludes that, since 

a* ; y = a ; x, 

and a, x are incommensurable in length, therefore a', y are incommensurable. 
But Euclid never commits the irregularity of proving a theorem by means of 
a later one. Gregory sought to get over the difficulty by putting x, 10 after 
x. ti ; but of course, if the order were so inverted, the Lemma would still be 
in the wrong place. 

Further, the expression Ifinffopfv yap, "for we have learnt (how to do this)," 
is not in Euclid's manner and betrays the hand of a learner (though the same 



X. io, n] PROPOSITIONS io, it 33 

expression is found in the Sectia Canenis of Euclid, where the reference is 
to the Elements). 

Lastly the manuscript P has the number io, in the first hand, at the top 
of x. it, from which it may perhaps be concluded that x. io had at first no 
number. 

It seems best therefore to reject as spurious both the Lemma and x. io. 

The argument of X. io is simple. If a be a given straight line and «, n 
numbers which have not to one another the ratio of square to square, take x 
such that 

a' :x' = m : n, [x. 6, Por.] 

whence a, x are incommensurable in length. [x, 9] 

Then take y a mean proportional between a, x, whence 

a* : y' = a ; x [v. Def. 9] 

[- Jm : J>t], 
and x is incommensurable in length only, white y is incommensurable in 
square as well as in length, with a. 



Proposition ii. 

If four magnitudes be proportional, and the first be com- 
mensurable with the second, the third will also be commensurable 
with the fourth ; and, if the first be incommensurable with the 
second, the third will also be incommensurable with the fourth. 

Let A y B, C, D be four magnitudes in proportion, so 
that, as A is to B, so is C 

to D, A b 

and let A be commensurable c D 

with B ; 

I say that C will also be commensurable with D. 

For, since A is commensurable with B, 
therefore A has to B the ratio which a number has to a 
number. [x. 5] 

And, as A is to B, so is C to D ; 
therefore C also has to D the ratio which a number has to a 
number ; 
therefore C is commensurable with D, [x. 6] 

Next, let A be incommensurable with B ; 
I say that C will also be incommensurable with D. 

For, since A is incommensurable with B, 
therefore A has not to B the ratio which a number has to a 
number. [x. 7] 



34 BOOK X [jc. ii, ii 

And, as A is to B, so is C to D ; 
therefore neither has C to D the ratio which a number has to 
a number ; 
therefore C is incommensurable with D. [x. 8] 

Therefore etc. 

I shall henceforth, for the sake of brevity, use symbols for the terms 
" commensurable (with) " and " incommensurable (with) " according to the 
varieties described in x. Deff. i — 4. The symbols are taken from Lorenz 
and seem convenient. 

Commensurable and commensurable with, in relation to areas, and com- 
mensurable in length and commensurable in length withy in relation to straight 
lines, will be denoted by <->, 

Commensurable in square only or commensurable in square only with (terms 
applicable only to straight lines) will be denoted by "-. 

Incommensurable {with), of areas, and incommensurable (with), of straight 
lines will be denoted by *j. 

Incommensurable in square (with) (a term applicable to straight lines only) 
will be denoted by <*-, 

Suppose a, b, c, d to be four magnitudes such that 
a : > m e i t, 

Then (1), if a ~ i, a : 6 = m : n, where m, n are integers, [x. 5] 

whence c \ d = m : tt, 

and therefore c « d. [x. 6] 

(j) If a u b, a :b*m : n, [x. 7] 

so that c:d*m:n, 

whence c u d, [x. 8] 



Proposition 12. 

Magnitudes commensurable with the same magnitude are 
commensurable with one another also. 

For let each of the magnitudes A, B be commensurable 
with C ; 
I say that A is also commensurable with B. 



-D 

— E H 



-F K 

-G — L 



For, gince A is commensurable with C, 
therefore A has to C the ratio which a number has to a 
number. [x. 5] 



x. is] PROPOSITIONS it, is 35 

Let it have the ratio which D has to E. 

Again, since C is commensurable with B, 

therefore C has to B the ratio which a number has to a 
number. [x. 5] 

Let it have the ratio which F has to G. 
And, given any number of ratios we please, namely the 
ratio which D has to E and that which F has to G, 

let the numbers H, K, L be taken continuously in the given 
ratios ; [cf. vin. 4] 

so that, as D is to E, so is H to K, 

and, as F is to G, so is K to L. 

Since, then, as A is to C, so is D to E, 
while, as D is to E, so is H to K, 
therefore also, as A is to C, so is H to K. ( [v. n] 

Again, since, as C is to 2?, so is 7*" to G, 
while, as /*" is to (r, so is K to Z., 
therefore also, as C is to B, so is A' to L. [v. 11] 

But also, as A is to C, so is // to K\ 
therefore, ex aequali, as A is to /?, so is H to Z. [v. «] 

Therefore -*4 has to B the ratio which a number has to a 
number ; 
therefore A is commensurable with B. [x. 6] 

Therefore etc. 

Q. E. D, 



We have merely to 

numbers. 


ge 


through the process of 


compounding 


two ratios in 


Suppose 
Therefore 




0, 6 each *"> c. 
a : c - m : 71, say, 




[*-s] 


Now 




e:b = P-g, say. 
m • h = tup : np, 






and 

Therefore 




p : g = «/ : nq. 
a ve=mp : «/, 






whence, ex aeqvati* 
so that 




c \ b = np : nq, 
a : b-mp : ruf, 
a * A, 




[*.«] 



36 BOOK X [x. 1 3l Lemma 

Proposition 13. 

If two magnitudes be commensurable, and the one of them 
be incommensurable with any magnitude, the remaining one 
will also be incommensurable with the same. 

Let A, B be two commensurable magnitudes, and let one 
of them, A, be incommensurable with 

any other magnitude C; A— 

I say that the remaining one, B, will 

also be incommensurable with C. & 

For, if B is commensurable with C, 
while A is also commensurable with B, 
A is also commensurable with C. [x. 1*] 

But it is also incommensurable with it : 
which is impossible. 

Therefore B is not commensurable with C\ 
therefore it is incommensurable with it. 

Therefore etc. 

L-j Erfrl MA. 

Given two unequal straight lines, to find by what square the 
square on the greater is greater than the square on the less. 

Let AB, C be the given two unequal straight lines, and 
let AB be the greater of them ; 
thus it is required to find by what 
square the square on AB is greater 
than the square on C. 

Let the semicircle ADB be de- 
scribed on AB, 

and let AD be fitted into it equal to C; [iv. 1] 

let DB be joined. 

It is then manifest that the angle ADB is right, [m. 31] 

and that the square on AB is greater than the square on 
AD, that is, C, by the square on DB. [i- 47] 

Similarly also, if two straight lines be given, the straight 
line the square on which is equal to the sum of the squares 
on them is found in this manner. 




Lemma, x. 14] 



PROPOSITIONS 13, 14 



37 



Let AD, DB be the given two straight lines, and let it be 
required to find the straight line the square on which is equal 
to the sum of the squares on them. 

Let them be placed so as to contain a right angle, that 
formed by AD, DB ; 
and let AB be joined. 

It is again manifest that the straight line the square on 
which is equal to the sum of the squares on AD, DB is AB. 

[•■ 47] 

Q. E. D. 

The lemma gives an obvious method of finding a straight line {e) equal to 
J<? t £*, where a, b are given straight lines of which a is the greater. 



Proposition 14. 

If four straight lines be proportional, and the square on 

the first be greater than the square on the second by the square 

on a straight line commensurable with the first, the square on 

the third will also be greater than the square on the fourth by 

5 the square on a straight line commensurable with the third. 

And, if the square on the first be greater than the square 
on the second by the square on a straight line incommensurable 
with the first, the square on the third will also be greater than 
t/ie square on the fourth by the square on a straight line in- 
to commensurable with the third. 

Let A, B, C, D be four straight lines in proportion, so 

that, as A is to B, so is C to D ; 

and let the square on A be greater than 

the square on B by the square on E, and 
islet the square on C be greater than the 

square on D by the square on F; 

I say that, if A is commensurable with E, 

C is also commensurable with F, 

and, if A is incommensurable with E, C is 
kj also incommensurable with F. 

For since, as A is to B, so is C to D, 

therefore also, as the square on A is to the square on B, so is 

the square on C to the square on D. [vi. 22] 

But the squares on E, B are equal to the square on A, 
15 and the squares on D, F are equal to the square on C. 



A B 



C D 



3« BOOK X [x. 14 

Therefore, as the squares on E, B are to the square on 
B, so are the squares on D, F to the square on D ; 

therefore, separando, as the square on E is to the square on 
B, so is the square on F to the square on D ; [v. 17] 

30 therefore also, as E is to B, so is F to D ; [vi. 22] 

therefore, inversely, as B is to E, so is D to F. 

But, as v4 is to B, so also is C to D \ 

therefore, ex aequali, as A is to .£", so is C to T 7 ! [v. ta] 

Therefore, if ^4 is commensurable with ZT, C is also com- 
35 mensurable with F, 

and, if A is incommensurable with E, C is also incommen- 
surable with F. fx. iil 

Therefore etc. 

3, 5, 8, 1o< Euclid speaks of the square on the lint {third) being greater thin the square 
on the second {fourth) by the square on a straight line commensurable {incommensurable) 
"with if*,//' tiat'Tri),'* and similarly in all Like phrases throughout the Book, For clearness 
sake I substitute "the first," " the third," or whatever it may be, for " itself" in these cases. 

Suppose a, i, c, d to be straight lines such that 

a : b = c ;d , /i). 

ft follows [vt. u] that d> :P = c'':d 1 (2). 

In order to prove that, eenvcrtendo, 

a' : (a i -P) = £* : (S-d*) 

Euclid has to use a somewhat roundabout method owing to the absence of a 
coimertcnda proposition in his Book v. (which omission Simson supplied by 
his Prop. E). 

It follows from (2) that 

{(*■-#>) + *»} :P^{tf-<P) + d*) :d\ 

whence, separando, (a 1 - **) ; #■= {/' -d 1 ) ; d\ [v. 17] 

and, inversely, P : (a* - P) = <T : (V - d*). 

From this and (2}, ex aeguali, 

a*: (a?-#)«^ : (c'-d*). [v, 22] 

Hence a : ■Ja*-P = e : -Jc' — d*. [vi. 22] 

According therefore as a«or u Ja' — P, 

e^oT^Jf-d'. [x. 11] 

If a « \/(7 J - P. we may put J a- - P = Aa, where k is of the form mjn 
and m, n are integers. And if \'a- - P = ka, it follows in this case that 
■fF-d* = kc. 



x. 15] PROPOSITIONS 14, 15 39 

Proposition 15. 

If two commensurable magnitudes be added together, the 
whole will also be commensurable with each of them / and, if 
the whole be commensurable with one of them, the original 
magnitudes will also be commensurable. 

For let the two commensurable magnitudes AB, BC be 
added together ; 

I say that the whole AC is also A ■ c 

commensurable with each of the 
magnitudes AB, BC. 

For, since AB, BC are commensurable, some magnitude 
will measure them. 

Let it measure them, and let it be D. 

Since then D measures AB, BC, it will also measure the 
whole AC. 

But it measures AB, BC also ; 
therefore D measures AB, BC, AC ; 

therefore AC is commensurable with each of the magnitudes 
AB, BC. [x. Def. 1] 

Next, let AC be commensurable with AB; 
I say that AB, BC are also commensurable. 

For, since AC, AB axe commensurable, some magnitude 
will measure them. 

Let it measure them, and let it be D. 

Since then D measures CA, AB, it will also measure the 
remainder BC. 

But it measures AB also ; 

therefore D will measure AB, BC; 

therefore AB, BC are commensurable, [x. Def. r] 

Therefore etc. 

(1) If a, b be any two commensurable magnitudes, they are of the form 
im, ne, where c is a common measure of a, b and m, n some integers. 

It follows that a + b = (m + n) c ; 

therefore (a + b), being measured by c, is commensurable with both a and /'. 

(2) If a + b is commensurable with either a or b, say a, we may put 
a + b = mc, a = nc, where c is a common measure of {a + b), a, and m, n are 
integers. 

Subtracting, we have b = («— ») t, 

whence /' *• a. 



4 o 



BOOK X [x. 16 



Proposition 16. 



If two incommensurable magnitudes be added together, the 
whole will also be incommensurable with each of them ; and, if 
the whole be incommensurable with one of them, the original 
magnitudes will also be incommensurable. 

For let the two incommensurable magnitudes AB, BC be 
added together ; 

I say that the whole AC is also incommensurable 
with each of the magnitudes AB, BC. 

For, if CA, AB are not incommensurable, some 
magnitude will measure them. 

Let it measure them, if possible, and let it be D. 

Since then D measures CA, AB, 

therefore it will also measure the remainder BC. 

But it measures AB also ; 
therefore D measures AB, BC. 

Therefore AB, BC are commensurable ; 
but they were also, by hypothesis, incommensurable : 
which is impossible. 

Therefore no magnitude will measure CA, AB ; 
therefore CA, AB are incommensurable. [x. Def. 1] 

Similarly we can prove that AC, CB are also incom- 
mensurable. 

Therefore ACh incommensurable with each of the magni- 
tudes AB, BC. 

Next, let AC be incommensurable with one of the magni- 
tudes AB, BC. 

First, let it be incommensurable with AB ; 
I say that AB, BC are also incommensurable. 

For, if they are commensurable, some magnitude will 
measure them. 

Let it measure them, and let it be D. 

Since then D measures AB, BC. 
therefore it will also measure the whole AC. 

But it measures AB also ; 
therefore D measures CA, AB, 



x. i6, 17] PROPOSITIONS 16, 17 41 

Therefore CA, AB are commensurable ; 
but they were also, by hypothesis, incommensurable : 
which is impossible. 

Therefore no magnitude will measure AB, BC ; 
therefore AB, BC are incommensurable. [x. Def. t] 

Therefore etc. 

Lemma. 

If to any straight line there be applied a parallelogram 
deficient by a square figure, the applied parallelogram is equal 
to the rectangle contained by the segments of the straight line 
resulting from the application. 

For let there be applied to the straight line AB the 
parallelogram AD deficient by the 
square figure DB; 

I say that AD is equal to the rectangle 
contained by AC, CB. 

This is indeed at once manifest ; 

for, since DB is a square, 

DC is equal to CB ; 

and AD is the rectangle AC, CD, that is, the rectangle AC, 
CB, 

Therefore etc. 

If a be the given straight line, and x the side of the square by which the 
applied rectangle is to be deficient, the rectangle is equal to ax- x*, which is 
of course equal to x {a - x). The rectangle may be written xy, where 
* + y = a. Given the area x(a-x), or xy (where x+y = a), two different 
applications will give rectangles equal to this area, the sides of the defect 
being x or a - x (x or y) respectively ; but the second mode of expression 
shows that the rectangles do not differ in form but only in position. 

Proposition 17. 

If there be two unequal straight lines, and to the greater 
there be applied a parallelogram equal to the fourth part of 
the square on the less and deficient by a square figure, and if 
it divide it into parts which are commensurable in length, then 
s the square on the greater will be greater than the square on 
the less by the square on a straight line commensurable with 
the greater. 

And, if the square on the greater be greater than the square 
on the less by the square on a straight line commensurable with 



4* BOOK X [x. 17 

10 the greater, and if there be applied to Ike greater a parallelogram 
equal to the fourth part of the square on the less and deficient 
by a square figure, it will divide it into parts which are com- 
mensurable in length. 

Let A, BC be two unequal straight lines, of which BC is 

15 the greater, 

and let there be applied to BCs. parallel- % 

ogram equal to the fourth part of the 

square on the less, A, that is, equal to 

the square on the half of A, and deficient 
to by a square figure. Let this be the k f e o 6 

rectangle BD, DC, [cf. Lemma] 

and let BD be commensurable in length with DC; 

I say that the square on BC is greater than the square on A 

by the square on a straight line commensurable with BC. 
*5 For let BC be bisected at the point E, 

and let EF be made equal to DE. 

Therefore the remainder DC is equal to BF. 

And, since the straight line BC has been cut into equal 

parts at E, and into unequal parts at D, 

30 therefore the rectangle contained by BD, DC, together with 
the square on ED, is equal to the square on EC ; [n. 5] 

And the same is true of their quadruples ; 
therefore four times the rectangle BD, DC, together with 
four times the square on DE, is equal to four times the square 
35 on EC. 

But the square on A is equal to four times the rectangle 
BD, DC; 

and the square on DF is equal to four times the square on 
DE, for DF is double of DE. 
40 And the square on BC is equal to four times the square 
on EC, for again BC is double of CE. 

Therefore the squares on A, DE are equal to the square 
on BC, 

so that the square on BC is greater than the square on A by 
45 the square on DF. 

It is to be proved that BC is also commensurable with DF. 
Since BD is commensurable in length with DC, 
therefore BC is also commensurable in length with CD. [x. 15] 



x. 17] PROPOSITION 17 43 

But CD is commensurable in length with CD, BF, for 

so CD is equal to BF. [x. 6] 

Therefore BC is also commensurable in length with BF, 

CD, [x. 12] 

so that BC is also commensurable in length with the remainder 
FD; [x. .5] 

55 therefore the square on BC is greater than the square on A 
by the square on a straight line commensurable with BC. 

Next, let the square on BC be greater than the square on 
A by the square on a straight line commensurable with BC, 
let a parallelogram be applied to BC equal to the fourth part 
60 of the square on A and deficient by a square figure, and let 
it be the rectangle BD, DC. 

It is to be proved that BD is commensurable in length 
with DC. 

With the same construction, we can prove similarly that 
65 the square on BC is greater than the square on A by the 
square on FD. 

But the square on BC is greater than the square on A 
by the square on a straight line commensurable with BC. 
Therefore BC is commensurable in length with FD, 
70 so that BC is also commensurable in length with the remainder, 
the sum of BF, DC. [x. 15] 

Bat the sum of BF, DC is commensurable with DC, [x. 6] 
so that BC is also commensurable in length with CD \ [x. 12] 

and therefore, separando, BD is commensurable in length 
n with DC. [x 15] 

Therefore etc. 

15- After raying literally that "the square on BC is greater than the square on A by the 
square on DF" Euclid adds the equivalent expression with &vhvf*l in its technical sense, 
it UP apa Tiji A a<.t{w &6yarat t£ AZ. As this is untranslatable in English except by a 
paraphrase in practically the< same words as have preceded, 1 have not attempted to 
reproduce it. 

This proposition gives the condition that the roots of the equation in x, 
ax-x* = p(= -, sayV 

are commensurable with a, or that x is expressible in terms of a and integral 

numbers, i.e. is of the form — a. No better proof can be found for the fact 

that Euclid and the Greeks used their solutions of quadratic equations for 
ntoaerica! problems. On no other assumption could an elaborate discussion 
of the conditions of incommensurability of the roots with given lengths o: 



44 BOOK X [x. 17 

with a given number of units of length be explained. In a purely geometrical 
solution the distinction between commensurable and incommensurable roots 
has no point, because each can equally easily be represented by straight lines. 
On the other hand, on the assumption that the numerical solution of quadratic 
equations was an important part of the system of the Greek geometers, 
the distinction between the cases where the roots are commensurable and 
incommensurable respectively with a given length or unit becomes of great 
importance. Since the Greeks had no means of expressing what we call an 
irrational number, the case of an equation with incommensurable roots could 
only be represented by them geometrically; and the geometrical representations 
had to serve instead of what we can express by formulae involving surds, 

Euclid proves in this proposition and the next that, x being determined 
from the equation 

x(a-x) = - (i\ 

4 

x, (a~x) are commensurable in length when Ja' — P, tf are so, and incom- 
mensurable in length when Va* — P,a are incommensurable ; and conversely. 
Observe the similarity of his proof to our algebraical method of solving 
the equation, a being represented in the figure by BC, and x by CD, 

EF=ED=°~x 

2 

— x\ - — , by Eucl. u. 5. 

If we multiply throughout by 4, 

whence, bv (1), & + (a — txf = a', 

or J-P=(a-2xY, 

and ■Ja , -6* = a-ix. 

We have to prove in this proposition ^^^_ 

(1) that, if x, {a — x) are commensurable in length, so are a, J a' — P, 
(a) that, if a, >/a* - *" are commensurable in length, so are x, {a — x). 

(1) To prove that a, a- 2X are commensurable in length Euclid employs 
several successive steps, thus. 

Since (a - x) « x, a « x. 

But x « 2X. 

Therefore a » »x 

« (a - »x). 
That is, a « J*? -P. 

(2) Since a « */a* — #", a " a- ix, 
whence " " '*■ 

But 2X»x; 

therefore * ** x, 

and hence (*-•*) " * 



[X. 


•5] 


[X 


.«] 


[X. 


n] 


[X. 


■5] 


[X. 


■5] 


[* 


.61 


[X. 


12] 


[X. 


'5] 



x. 17, 1 8] PROPOSITIONS 17, 1 8 45 

It is often more convenient to use the symmetrica.) form of equation in 

this and similar cases, viz. 

x + y = a j 

The result with this mode of expression is that 

(1) if x " y, then a « nJa'-P; and 

(3) if a « iJj—&, then x * y. 

The truth of the proposition is even easier to see in this case, since 
( X -yf = (<?-?). 



Proposition 18. 

If there be two unequal straight lines, and to the greater 
there be applied a parallelogram equal to the fourth part of 
the square on the less and deficient by a square figure, and 
if it divide it into parts which are incommensurable, the square 
on the greater will be greater than the square on the less by 
the square on a straight line incommensurable with the greater. 

And, if the square on the greater be greater titan the square 
on the less by the square on a straight line incommensurable 
with the greater, and if there be applied to the greater a 
parallelogram equal to the fourth part of the square on the 
less and deficient by a square figure, it divides it into parts 
which are incommensurable. 

Let A, BC be two unequal straight lines, of which BC is 
the greater, 

and to BC let there be applied a parallelogram equal 
to the fourth part of the square on the less, A, and 
deficient by a square figure. Let this be the rect- 
angle BD, DC, [cf. Lemma before x. 17] 
and let BD be incommensurable in length with DC; 
I say that the square on BC is greater than the 
square on A by the square on a straight line incom- 
mensurable with BC. 

For, with the same construction as before, we can prove 
similarly that the square on BC is greater than the square on 
A by the square on FD. 

It is to be proved that BC is incommensurable in length 
with DF. 



46 BOOK X [x. 18 

Since BD is incommensurable in length with DC, 

therefore BC is also incommensurable in length with CD. 

[x. .6] 

But DC is commensurable with the sum of BF, DC ; [x. 6] 
therefore BC is also incommensurable with the sum of BF, 
DC; [x. I3 ) 

so that.5C is also incommensurable in length with the remainder 
FD. [x. 16] 

And the square on BC is greater than the square on A 
by the square on FD \ 

therefore the square on BC is greater than the square on A 
by the square on a straight line incommensurable with BC. 

Again, let the square on BC be greater than the square on 
A by the square on a straight line incommensurable with BC, 
and let there be applied to BC a parallelogram equal to the 
fourth part of the square on A and deficient by a square figure. 
Let this be the rectangle BD, DC. 

It is to be proved that BD is incommensurable in length 
with DC. 

For, with the same construction, we can prove similarly 
that the square on BC is greater than the square on A by 
the square on FD. 

But the square on BC is greater than the square on A by 
the square on a straight line incommensurable with BC; 
therefore BC is incommensurable in length with FD, 
so that BC is also incommensurable with the remainder, the 
sum of BF, DC, [x. 16] 

But the sum of BF, DC is commensurable in length with 
DC; [x. 6] 

therefore BC is also incommensurable in length with DC, 

[x. n\ 
so that, separando, BD is also incommensurable in length with 
DC. [x. 16] 

Therefore etc. 

With the same notation as before, we have to prove in this proposition that 
(1) if (a - x), x a re incommensurable in length, so are a, J a' - ?, and 
(a) if a, V«" — P are incommensurable in length, so are (a - x), x. 

Or, with the equations 

P\ 
xy = - 
4 > 
x +y = a 



x. i8, 19] PROPOSITIONS 18, 19 47 

(1) if x -jy, then a w Ja' — P, and 

(2) if a 1/ Va* - #", then .* ^ j>. 

The steps are exactly the same as shown under (1) and (2) of the last 
note, with u instead of n , except only in the lines "x 'ft zx" and "m " x" 
which are unaltered, while, in the references, x. 13, 16 take the place of x. 
12, 15 respectively. 

[Lemma. 

Since it has been proved that straight lines commen- 
surable in length are always commensurable in square also, 
while those commensurable in square are not always com- 
mensurable in length also, but can of course be either 
commensurable or incommensurable in length, it is manifest 
that, if any straight line be commensurable in length with a 
given rational straight line, it is called rational and commen- 
surable with the other not only in length but in square also, 
since straight lines commensurable in length are always 
commensurable in square also. 

But, if any straight line be commensurable in square with 
a given rational straight line, then, if it is also commensurable 
in length with it, it is called in this case also rational and 
commensurable with it both in length and in square ; but, if 
again any straight line, being commensurable in square with a 
given rational straight line, be incommensurable in length 
with it, it is called in this case also rational but commensurable 
in square only.] 

Proposition 19. 

The rectangle contained by rational straight lines commen- 
surable in length is rational. 

For let the rectangle AC be contained by the rational 
straight lines AB, BC commensurable in 
length ; 
I say that AC is rational. 

For on AB let the square AD be de- 
scribed ; 
therefore AD is rational. [x. Def. 4] 

And, since AB is commensurable in 
length with BC, 
while AB is equal to BD, 
therefore BD is commensurable in length with BC. 




4 8 BOOK X [x. i 9 

And, as BD is to BC, so is DA to AC. [vi. i] 

Therefore ZM is commensurable with /4C. [x. iij 

But DA is rational ; 
therefore AC is also rational. [x. Def, 4] 

Therefore etc. 

There is a difficulty in the text of the enunciation of this proposition. 

The Greek runs to vwq pifnav fXTjKti &vp,p£Tp*iiv Kara T*pa totv TrpouprfpAvrnv 
Tpoirwv tv9tifj>v Tripitxpfifvov op6oywvior prjrav «rrcy, where the rectangle is 
said to be contained by " rational straight lines commensurable in length in 
any of the aforesaid ways." Now straight lines can only be commensurable 
in length in one way, the degrees of commensurability being commensurability 
in length and commensurability in square only. But a straight tine may be 
rational in two ways in relation to a given rational straight line, since it may 
be either commensurable in length, or commensurable in square only, with the 
latter. Hence Billingsley takes Kara rtva Tajf irpofiprfitaitiiv jporuv with ptpw, 
translating " straight lines commensurable in length and rational in any of the 
aforesaid ways," and this agrees with the expression in the next proposition 
"a straight line once more rational in any of the aforesaid ways"; but the 
order of words in the Greek seems to be fatal to this way of translating 
the passage. 

The best solution of the difficulty seems to be to reject the words "in 
any of the aforesaid ways " altogether. They have reference to the I»emma 
which immediately precedes and which is itself open to the gravest suspicion. 
It is very prolix, and cannot be called necessary ; it appears moreover in 
connexion with an addition clearly spurious and therefore relegated by 
Heiberg to the Appendix. The addition does not even pretend to be Euclid's, 
for it begins with the words "for he calls rational straight lines those — " 
Hence we should no doubt relegate the Lemma itself to the Appendix. 
August does so and leaves out the suspected words in the enunciation, as I 
have done. 

Exactly the same arguments apply to the Lemma added (without the 
heading " Lemma ") to x. 33 and the same words " in any of the aforesaid 
ways " used with " medial straight lines commensurable in length " in the 
enunciation of x. 24. The said Lemma must stand or fall with that riow in 
question, since it refers to it in terms: "And in the same way as was explained 
in the case of rational*.. .." 

Hence I have bracketed the Lemma added to x. 23 and left out the 
objectionable words in the enunciation of x. 24. 

If p be one of the given rational straight lines {rational of course in the 
sense of x. Def. 3), the other can be denoted by kp, where * is, as usual, of 
the form mjn (where jw, « are integers). Thus the rectangle is kp\ which is 
obviously rational since it is commensurable with p'. [x. Def. 4.] 

A rational rectangle may have any of the forms ab, ha", kA or A, where 
a, b are commensurable with the unit of length, and A with the unit of area. 

Since Euclid is not able to use ip as a symbol for a straight line 
commensurable in length with p, he has to put his proof in a form corre- 
sponding to 

p ] : V = P ' *P. 
whence, p, kp being commensurable, p 5 , hp- are so also. [x. 11] 




x-io, 2i] PROPOSITIONS 19—21 49 



Proposition 20. 

If a rational area be applied to a rational straight line, it 
produces as breadth a straight line rational and commensurable 
in length with the straight line to which it is applied. 

For let the rational area AC he. applied to AB, a straight 
line once more rational in any of the aforesaid 
ways, producing BC as breadth ; 
I say that BC is rational and commensurable in 
length with BA. 

For on AB let the square AD be described ; 
therefore AD is rational. [x. Def. 4] 

But AC is also rational ; 
therefore DA is commensurable with AC. 

And, as DA is to AC, so is DB to BC. 

[v.. il 

Therefore DB is also commensurable with BC ; [x. n] 

and DB is equal to BA ; 

therefore AB is also commensurable with BC. 

But AB is rational ; 
therefore BC is also rational and commensurable in length 
with AB. 

Therefore etc. 

The converse of the last. If p is a rational straight line, any rational area 
is of the form kp\ If this be "applied" to p, the breadth is kp commensurable 
in length with p and therefore rational. We should reach the same result if 
we applied the area to another rational straight line <r. The breadth is then 

V V m ,, 

— = ~ ,tr = — k .a QT fc<r, say. 



Proposition 21. 

The rectangle contained by rational straight lines commen- 
surable in square only is irrational and the side of the square 
equal to it is irrational. Let the latter be called medial. 

For let the rectangle AC be contained by the rational 
straight lines AB, BC commensurable in square only ; 



S° 



BOOK X 



[X. 31 



I say that AC is irrational, and the side of the square equal 

to it is irrational ; 

and let the latter be called medial. 

For on AB let the square AD be described ; 
therefore AD is rational. [x. Def. 4] 

And, since AB is incommensurable in length 
with BC, 

for by hypothesis they are commensurable in 
square only, 
while AB is equal to BD, 
therefore DB is also incommensurable in length with BC. 

And, as DB is to BC, so is AD to AC; [n, 1] 

therefore DA is incommensurable with AC. [x. it] 

But DA is rational ; 
therefore AC is irrational, 

so that the side of the square equal to AC is also irrational. 

[\. Def. 4] 
Ana Jet the latter be called medial. 

Q. E. D. 

A medial straight line, now defined for the first time, is so called because 
it is a mean proportional between two rational straight lines commensurable 
in square only. Such straight lines can lie denoted by p, p ,J&. A medial 

straight line is therefore of the form Jp r jk or £*p. Kuclid's proof that this is 
irrational is equivalent to the following. Take p, p,JA commensurable in 
square only, so that they are incommensurable in length. 

Now p : p*]k - p' : p'^k, 

whence [x. 11] p'^ i is incommensurable with p 1 and therefore irrational 
[x. Def. 4], so that JpTJk is also irrational [#&.]. 

A medial straight line may evidently take either of the forms JaJB or 
XI AB, where of course B is not of the form k'A. 

Lemma, 

If there be two straight lines, then, as the first is to the 
second, so is the square on the first 
to the rectangle contained by the 
two straight lines. 

Let FE, EG be two straight 
lines. 

I say that, as FE is to EG, so is the square on FE to 
the rectangle FE, EG. 



X. xi, ti] 



PROPOSITIONS ai, 22 



5' 



For on FE let the square DF be described, 
and let GD be completed. 

Since then, as FE is to EG, so is FD to DG, [vi. i] 

and FD is the square on FE, 

and DG the rectangle DE, EG, that is, the rectangle FE, EG, 
therefore, as FE is to EG, so is the square on FE to the 
rectangle FE, EG. 

Similarly also, as the rectangle GE, EF is to the square 
on EF, that is, as GD is to FD, so is GE to EF. 

Q. E. D. 

If a, b be two straight lines, 

a : h - a* : ab. 



Proposition 22. 

The square on a medial straight line, if applied to a 
rational straight line, produces as breadth a straight line 
rational and incommensurable in length with that to which it 
is applied. 

Let A be medial and CB rational, 
and let a rectangular area BD equal to the square on A be 
applied to BC, producing CD as 
breadth ; 

I say that CD is rational and incom- 
mensurable in length with CB. 

For, since A is medial, the square 
on it is equal to a rectangular area 
contained by rational straight lines 
commensurable in square only. 

[x 21] 

Let the square on it be equal to GF. 

But the square on it is also equal to BD ; 
therefore BD is equal to GF. 

But it is also equiangular with it ; 
and in equal and equiangular parallelograms the sides about 
the e^ual angles are reciprocally proportional ; [vi. 14) 

therefore, proportionally, as BC is to EG, so is EF to CD. 

Therefore also, as the square on BC is to the square on 
EG, so is the square on EF to the square on CD. [vi. 22] 



B 




C 










< 




> 1 




J 1 


i F 



ja BOOK X [x. i* 

But the square on CB is commensurable with the square 
on EG, for each of these straight lines is rational ; 

therefore the square on EF is also commensurable with the 
square on CD. [x. n] 

But the square on EF is rational ; 
therefore the square on CD is also rational j [x. Uef. 4J 

therefore CD is rational. 

And, since EF is incommensurable in length with EG, 

for they are commensurable in square only, 

and, as EF is to EG, so is the square on EF to the rectangle 
FE, EG, [Lemma] 

therefore the square on EF is incommensurable with the 
rectangle FE, EG. [x. 11] 

But the square on CD is commensurable with the square 
on EF, for the straight lines are rational in square ; 
and the rectangle DC, CB is commensurable with the rect- 
angle FE, EG, for they are equal to the square on A j 
therefore the square on CD is also incommensurable with the 
rectangle DC, CB. [x. 13] 

But, as the square on CD is to the rectangle DC, CB, so 
is DC to CB ; [Umma] 

therefore DC is incommensurable in length with CB. [x. 1 1] 

Therefore CD is rational and incommensurable in length 
with CB. 

Q. E. D. 

Our algebraical notation makes the result of this proposition almost self- 
evident. We have seen that the square of a medial straight line is of the form 
J A . p 1 . If we "apply" this area to another rational straight line <r, tin- 
breadth is . 

This is equal to ' . o- = Jk . — <r, where m, n are integers. The latter 
<r ft 

straight line, which we may express, if we please, in the form Jk' . <r, is clearly 
commensurable with 0- in square only, and therefore rational but incom- 
mensurable in length with tr. 

Euclid's proof, necessarily longer, is in two parts. 

Suppose that the rectangle fo. p*~tr.x. 

Then (r) <r ;p= Jk.p:x, [vi. 14] 

whence o- 1 : p' = kf? ; Jr*. [vi. «] 

But <r* " p\ and therefore V « a*. [x. 1 1] 



X. 22, 13] 



PROPOSITIONS 22, 23 



And kp* is rational , 
therefore x*, and therefore x, is rational. 

(2) Since JJk . p < — p, ,Jk . p v p. 

But [Lemma] Jk . p : p = kp* : Jt, p 1 , 

whence ip» ^ Jk . p'. 

But Jk.p' = trx, and ip' « ;*■* (from above) ; 
therefore x**/<rx ; 

and, since «" : trx = x : o-, 

X ^ «r. 



53 
[x. Def. 4) 

[x. II] 
[Lemma] 



O F 



Proposition 23, 

W straight line commensurable with a medial straight line 
is medial. 

Let A be medial, and let B be commensurable with A ; 
I say that B is also medial. 

For let a rational straight line CD 

be set out, 

and to CD let the rectangular area CE 
equal to the square on A be applied, 
producing ED as breadth ; 
therefore ED is rational and incommen- 
surable in length with CD. [x. 22] 

And let the rectangular area CF 
equal to the square on B be applied to 
CD, producing DF as breadth. 

Since then A is commensurable with B, 
the square on A is also commensurable with the square on B. 

But EC is equal to the square on A, 
and CF is equal to the square on B ; 
therefore EC is commensurable with CF. 

And, as EC is to CF, so is ED to DF; fvi. 1] 

therefore .£"/? is commensurable in length with DF. [x. n] 

But ED is rational and incommensurable in length with 
DC; 

therefore DF is also rational [x. Def. 3] and incommensurable 
in length with DC. [x- 13] 

Therefore CD, DF are rational and commensurable in 
square only. 



54 BOOK X [x. 33 

But the straight line the square on which is equal to the 
rectangle contained by rational straight lines commensurable 
in square only is medial ; [x. 21] 

therefore the side of the square equal to the rectangle CD, 
DB"is medial. 

And B is the side of the square equal to the rectangle 
CD, DF; 
therefore B is medial. 

Porism. From this it is manifest that an area commen- 
surable with a medial area is medial. 

[And in the same way as was explained in the case of 
rationals [Lemma following x. 18] it follows, as regards medials, 
that a straight line commensurable in length with a medial 
straight line is called medial and commensurable with it not 
only in length but in square also, since, in general, straight 
lines commensurable in length are always commensurable in 
square also. 

But, if any straight line be commensurable in square with 
a medial straight line, then, if it is also commensurable in 
length with it, the straight lines are called, in this case too, 
medial and commensurable in length and m square, but, if in 
square only, they are called medial straight lines commen- 
surable in square only.] 

As explained in the bracketed passage following this proposition, a straight 
tine commensurable with a medial straight line in square only, as well as a 
straight line commensurable with it in length, is medial. 

Algebraical notation shows this easily. 

If i*p be the given straight line, Xi'p is a straight line commensurable 
in length with it and Jk . jb*p a straight line commensurable with it in square 
only. 

But if and Jk.p are both rational [x. Def. 3] and therefore can be 

expressed by p', and we thus arrive at &*p, which is clearly medial. 

Euclid's proof amounts to the following. 

Apply both the areas J&-p' and k'Ji.p' (or k^/Jk.p r ) to a rational 
straight line a. 

The breadths Jk . - and k'Ji . - f or kji . — I are in the ratio of the 

areas *Jk.p* and k*Jk . p 1 (or kji.p 1 ) themselves and are therefore com- 
mensurable. 

Now [x. 22] y/i . — is rational but incommensurable with <r. 
Therefore A»^£ . " for k^/Jt . —I is so also; 



x. 23. 24] PROPOSITIONS 33, 14 5S 

whence the area X'^/k , />* (or a^/£ . p') is contained by two rational straight 
lines com mensurable in square only, so that \k*p (or ^/\ , k*p) is a medial 
straight line. 

It is in the Porism that we have the first mention of a medial area. It is 
the area which is equal to the square on a medial straight line, an area, there- 
fore, of the form Iflf?, which !s, as a matter of fact, arrived at, though not 
named, before the medial straight line itself (x. 21). 

The Porism states that Ai'p s is a medial area, which is indeed obvious. 

Proposition 24. 

The rectangle contained by medial straight lines commen- 
surable in length is medial. 

For let the rectangle AC be contained by the medial 
straight lines AB, BC which are commensurable 
in length ; 
I say that AC is medial. 

For on AB let the square AD be described ; 
therefore AD is medial. 

And, since AB is commensurable in length 
with BC, 

while AB is equal to BD, 
therefore DB is also commensurable in length 
with BC; 
so that DA is also commensurable with AC, [vi. 1, x. 11] 

But DA is medial ; 
therefore AC is also medial. [x. 23, Por.] 

Q. E. D. 

There is the same difficulty in the text of this enunciation as in that of 
X. 19. The Greek says " medial straight lines commensurable in length in 
any of the aforesaid ways"; but straight lines can only be commensurable in 
length in one way, though they can be medial in two ways, as explained in the 
addition to the preceding proposition, i.e. they can be either commensurable 
in length or commensurable in square only with a given medial straight line. 
For the same reason as that explained in the note on x. 19 I have omitted 
"in any of the aforesaid ways " in the enunciation and bracketed the addition 
to x. 33 to which it refers. 

k*p and \lrp are medial straight lines commensurable in length. The 

rectangle contained by them is \k*p\ which may be written &*p"' and is there- 
fore clearly medial. 

Euclid's proof proceeds thus. Uti x, \x be the two medial straight lines 
commensurable in length. 

Therefore «* : x . \x = x : Xx. 



56 



BOOK X 






[x. 24, 25 
[x. ,«] 



But x « Ajt, so that x 1 «• x . Ajc, 

Now x? is medial (x. a i ] ; 
therefore x i Kx is also medial. [x. 23, For.) 

We may of course write two medial straight lines commensurable in length 
in the forms m/irp, ntrp; and these may either be mJaJB, nJa^JB, or 
mtlAB, ni/AB. 



Proposition 25. 

rke rectangle contained by medial straight lines commen- 
surable in square only is either rational or medial. 

For let the rectangle AC be contained by the medial 
straight lines AB, BC which are 
commensurable in square only ; 
1 say that AC is either rational 
or medial. 

For on AB, BC let the 
squares AD, BE be described ; 
therefore each of the squares 
AD, BE is medial. 

Let a rational straight line 
FG be set out, 
to FG let there be applied the rectangular parallelogram GH 
equal to AD, producing FH as breadth, 
to HM let there be applied the rectangular parallelogram MK 
equal to AC, producing HK as breadth, 
and further to K N let there be similarly applied NL equal to 
BE, producing KL as breadth ; 
therefore FH, HK, KL are in a straight line. 

Since then each of the squares AD, BE is medial, 
and AD is equal to GH, and BE to NL, 
therefore each of the rectangles GH, ML is also medial. 

And they are applied to the rational straight lino FG ; 
therefore each of the straight lines FH, KL is rational and 







\ 





H 


G 






M 


D 


B 






K 


N 




( 


> E 






L 



incommensurable in length with FG. 

And, since AD is commensurable with BE, 
therefore GH is also commensurable with NL. 

And, as GH is to NL, so is FH to KL ; 
therefore FH is commensurable in length with KL. 



[x. 2 2] 



[v,. ,] 

[x 1 ,] 



x. 25] PROPOSITIONS 24, 25 57 

Therefore FH, KL are rational straight lines commen- 
surable in length ; 
therefore the rectangle FH, KL is rational. [x. 19] 

And, since DB is equal to BA, and OB to BC, 
therefore, as DB is to BC, so is AB to BO. 

But, as DB is to BC, so is DA to AC, [vt. 1] 

and, as AB is to BO, so is AC to CO ; [»rf.] 

therefore, as DA is to A C, sots AC to CO. 

But AD is equal to GH, AC to J/A" and CO to #Z ; 
therefore, as GH is to A/A', so is MK to AX ; 
therefore also, as FH is to HK, so is HK to A'Z. ; [vi. 1, v. n] 
therefore the rectangle FH, KL is equal to the square on HK. 

fvi. 17] 

But the rectangle FH, KL is rational ; 
therefore the square on HK is also rational. 

Therefore HK is rational. 

And, if it is commensurable in length with FG, 

HN is rational ; [*. 19] 

but, if it is incommensurable in length with FG, 

KH, HM atz rational straight lines commensurable in square 
only, and therefore HN is medial. [x. 21] 

Therefore HN is either rational or medial. 

But HN is equal to AC; 

therefore AC is either rational or medial. 

Therefore etc. 

Two medial straight lines commensurable in square only are of the form 

The rectangle contained by them is ,/A ■ *V*. Now this is in general 
medial; but, if N /X = k' Jk, the rectangle is kk'/?, which is rational. 

Euclid's argument is as follows. Let us, for convenience, put x for frp, sc 
that the medial straight lines are x, J\ . x. 

Form the areas x*, x . J\ . x, Xx*, 
and let these be respectively equal to <ru, <ru, mv, where a is a rational 
straight line. 

Since X s , Kx' are medial areas, 
so are av, <rw, 
whence v, w are respectively rational and <*- <r. 



5« 



BOOK X 



[x. *S. *<> 



But a? n kx 1 , 
so that o-B " oro, 
or u * w 

Therefore, u, w being both rational, «w is rational 

Now *■; J\.a* = J\.3*;\& 

Or <M:trv = tTV: trw, 

so that u:v = v:w, 

and arc = p". 

Hence, by {2), »>, and therefore v, is rational 

Now (a) if »<"» <r, (tv or ,/A . ^ is rational; 

(0) if I* v (r, so that v ^- <r, trv or J\ . x 1 is medial. 



,(,). 
.<i). 



■(3). 



Proposition 26. 
4 medial area does not exceed a medial area by a rational 



area. 




For, if possible, let the medial area AB exceed the medial 
area AC by the rational area 
DB, 

and let a rational straight line 
EF be set out ; 

to EF let there be applied the 
rectangular parallelogram FH 
equal to AB, producing EH as 
breadth, 

and let the rectangle FG equal to AC be subtracted ; 
therefore the remainder B> > is equal to the remainder KH. 

But DB is rational ; 
therefore KH is also rational. 

Since, then, each of the rectangles AB, AC is medial, 
and AB is equal to FH, and AC to FG, 
therefore each of the rectangles FH, FG is also medial. 

And they are applied to the rational straight line EF; 
therefore each of the straight lines HE, EG is rational and 
incommensurable in length with MF. [x. a 2] 

And, since \DB is rational and is equal to KH, 
therefore] KH is [also] rational ; 
and it is applied to the rational straight line EF; 



x. a6] PROPOSITIONS 25, 26 59 

therefore GH is rational and commensurable in length with 
EF. [x. 10] 

But EG is also rational, and is incommensurable in length 
with EF; 

therefore EG is incommensurable in length with GH. [x. 13] 
And, as EG is to GH, so is the square on EG to the 
rectangle EG, GH; 

therefore the square on EG is incommensurable with the 
rectangle EG, GH. [x. n] 

But the squares on EG, GH are commensurable with the 
square on EG, for both are rational ; 

and twice the rectangle EG, GH is commensurable with the 
rectangle EG, GH, for it is double of it ; [x. 6] 

therefore the squares on EG, GH are incommensurable with 
twice the rectangle EG, GH \ [x. 13] 

therefore also the sum of the squares on EG, GH and twice 
the rectangle EG, GH, that is, the square on EH [11. 4], is 
incommensurable with the squares on EG, GH. [x. 16] 

But the squares on EG, GH are rational ; 
therefore the square on EH is irrational. [x. Def. 4] 

Therefore EH is irrational. 

But it is also rational : 
which is impossible. 

Therefore etc. 

Q. E. D. 

"Apply" the two given medial areas to one and the same rational straight 
line p. They can then be written in the form p , k*p, p . X>. 

The difference is then taji- Jtyp' ; and the proposition asserts that this 
cannot be rational, i.e. (y/i- JA) cannot be equal to &'. Cf. the proposition 
corresponding to this in algebraical text-books. 

To make Euclid's proof clear we will put x for A*p and y for X'p. 

Suppose p (x -y) = pi, 

and, if possible, let pz be rational, so that a must be rational and " p ..(1). 

Since px, py are medial, 

x and y are respectively rational and v p (z ). 

From (t) and (2), y %j s. 

Now y%=y* :yz, 

so that y* ^tyz. 






6o 




BOOK X 


But 




/t2 , "/| 


and 




2JI2 1 


-.jw 


Therefore 




y + ^v tyf, 


whence 




{y + zf »{?*#), 


or 




X 1 «(/+**). 


And (/ + **) is 


rational ; 






therefore x 1 , and consequently 


x, is irrational. 


But, by (2), x is 


rational : 






which is impossible. 






Therefore ps is 


not rational. 
















[x. j 6, a 7 






Proposition 27. 

To find medial straight lines commensurable in square only 
which contain a rational rectangle. 

Let two rational straight lines A, B commensurable in 

square only be set out ; 

let C be taken a mean proportional between 
A, B, [v,. I3 ] 

and let it he contrived that, 

as A is to B, so is C to D. [vx 12'J 
Then, since A, B are rational and com- 
mensurable in square only, 

the rectangle A, B, that is, the square on C 
[vi 17], is medial. [x. 21] 

Therefore C is medial. [x. 21] 

And since, as A is to B, so is C to D, 
and A t B are commensurable in square only, 
therefore C, D are also commensurable in square only. [x. n] 

And C is medial ; 
therefore D is also medial. [x. 23, addition] 

Therefore C, D are medial and commensurable in square 
only, 

I say that they also contain a rational rectangle. 

For since, as A is to B, so is C to D, 
therefore, alternately, as A is to C, so is B to D. [v. 16] 

But, as A is to C, so is C to B ; 
therefore also, as C is to B, so is B to Z? ; 
therefore the rectangle C, Z? is equal to the square on B, 



x. a?. 28] PROPOSITIONS 26—28 61 

But the square on B is rational ; 
therefore the rectangle C, D is also rational. 

Therefore medial straight lines commensurable in square 
only have been found which contain a rational rectangle. 

Q. E. D. 

Euclid takes two rational straight lines commensurable in square only, say 
p, k*p. 

Find the mean proportional, i.e. i*p. 

Take x such that p : $p = $ p . x (,j_ 

This gives x - Irp, 
and the lines required are krp, A*p. 

For (a) &*p is medial. 

And (j8), by (1), since p «- £'p, 

4*p ~- J&P, 
whence [addition to x. 23], since &*p is media), 

it*p is also medial. 
The medial straight lines thus found may take either of the forms 

<■) -i^Js, J 23* or (2) Tab, J~^ a - 



Proposition 28. 



To find medial straight lines commensurable in square only 
which contain a medial rectangle. 

Let the rational straight lines A, B, C commensurable in 
square only be set out ; 

let D be taken a mean proportional between A, B, [vi, 13] 
and let it be contrived that, 

as B is to C, so is D to E. [vi. 12] 



Since A, B are rational straight lines commensurable in 
square only, 

therefore the rectangle A, B, that is, the square on D [vi. 17], 
is medial. [x. at) 



61 BOOK X [x. 18 

Therefore D is medial. [x. *i] 

And since B, C are commensurable in square only, 

and, as B is to C, so is D to E, 

therefore D, E are also commensurable in square only. fx. u] 
But Z> is medial ; 

therefore E is also medial. [x. 13, addition] 

Therefore D, E are medial straight lines commensurable 
in square only. 

I say next that they also contain a medial rectangle. 

For since, as B is to C, so is D to E, 
therefore, alternately, as B is to D, so is C to E. [v. 16] 

But, as B is to D, so is D to A ; 

therefore also, as Z? is to A, so is C to E ; 

therefore the rectangle A, C is equal to the rectangle D, E. 

[v.. ,6] 
But the rectangle A, C is medial ; [x. zr] 

therefore the rectangle D, E is also medial. 

Therefore medial straight lines commensurable in square 
only have been found which contain a medial rectangle. 

Q. E. D. 

Euclid takes three straight lines commensurable in square only, i.e. of the 
form p, ^Vi A ft and proceeds as follows. 

Take the mean proportional to p, A'p, i.e. £*p. 

Then take a: such that 

*V : \*p = t*p : x (r), 

so that x = A"p/A*. 

trp, A*p/£* are the required medial straight lines. 

For JPp is medial. 

Now, by (1), since t?p «- A*p, 

trp "- x, 

whence x is also medial [x. 33, addition], while «- £*p. 

Next, by (1), \*p : jc = **p : £*p 

= <i*p:p, 

whence x . A*p = A'p', which is medial. 

The straight lines Irp, X*p/£* of course take different forms according as 
the original straight lines are of the forms (1) a, ,JB, ,JC, (2) J A, JB, JC, 
(3) -J A, b, JC, and {4) J A, JB, e. 



x. a8, Lemma i] PROPOSITION 28 63 

E.g. in case (1) they are JaJB, <sj -jgi 



in case (2) they are VAB, */-— 









and so on. 

Lemma i. 

To find (wo square numbers such (hat their sum is also 
square. 

Let two numbers AB, BC be set out, and let them be 
either both even or both odd. 

Then since, whether an even a 6 S B 

number is subtracted from an 

even number, or an odd number from an odd number, the 
remainder is even, [ix. 24, 26] 

therefore the remainder AC is even. 

Let AC be bisected at D. 

Let AB, BC also be either similar plan„ numbers, or 
square numbers, which are themselves also similar plane 
numbers. 

Now the product of AB, BC together with the square on 
CD is equal to the square on BD. [11. 6] 

And the product of AB, BC is square, inasmuch as it 
was proved that, if two similar plane numbers by multiplying 
one another make some number the product is square, [tx. 1] 

Therefore two square numbers, the product of AB, BC, 
and the square on CD, have been found which, when added 
together, make the square on BD. 

And it is manifest that two square numbers, the square 
on BD and the square on CD, have again been found such 
that their difference, the product of AB, BC, is a square, 
whenever AB, BC are similar plane numbers. 

But when they are not similar plane numbers, two square 
numbers, the square on BD and the square on DC, have been 
found such that their difference, the product of AB, BC, is 
not square. 

Q. E. D. 

Euclid's method of forming right-angled triangles in integral numbers. 
already alluded to in the note on 1. 47, is as follows. 

Take two similar plane numbers, e.g. mn/', mnq\ which are either both even 
or both odd, so that their difference is divisible by 1. 



64 BOOK X [Lemmas i, i 

Now the product of the two numbers, or wiW^'f 3 , is square, [ix, i] 

and, by n. 6, 

»,nf . mtf + (*»?---*»?)' = (^WJ, 

so that the numbers mnpq, £ (mitj? - mnq*) satisfy the condition that the sum 
of their squares is also a square number. 

It is also clear that \ (mnp* -t mnij*), mnpq are numbers such that the 
difftrcnce of their squares is also square. 



Lemma 2. 

To find two square numbers such that their sum is not 
square. 

For let the product of AB, BC, as we said, be square, 
and CA even, 
and let CA be bisected by D. 

E 

a <£ A 6 ' p ~~c" b 

It is then manifest that the square product of AB, BC 
together with the square on CD is equal to the square on BD. 

[See I-enima 1] 

Let the unit DE be subtracted ; 
therefore the product of AB, BC together with the square on 
CE is less than the square on BD. 

I say then that the square product of AB, BC together 
with the square on CE will not be square. 

For, if it is square, it is either equal to the square on BE, 
or less than the square on BE, but cannot any more be 
greater, lest the unit be divided. 

First, if possible, let the product of AB, BC together 
with the square on CE be equal to the square on BE, 
and let GA be double of the unit DE. 

Since then the whole AC is double of the whole CD, 
and in them AG is double of DE, 

therefore the remainder GC is also double of the remainder EC; 
therefore GC is bisected by E. 

Therefore the product of GB, BC together with the square 
on CE is equal to the square on BE. [it. 6] 

But the product of AB, BC together with the square on 
CE is also, by hypothesis, equal to the square on BE ; 



Lemma a] LEMMAS TO PROPOSITIONS 19, 30 fi 5 

therefore the product of GB, BC together with the square on 
CE is equal to the product of AB, BC together with the 
square on CE. 

And, if the common square on CE be subtracted, 
it follows that AB is equal to GB : 
which is absurd. 

Therefore the product of AB, BC together with the square 
on CE is not equal to the square on BE. 

I say next that neither is it less than the square on BE. 

For, if possible, let it be equal to the square on BE, 
and let HA be double of DF. 

Now it will again follow that HC is double of CF; 
so that CH has also been bisected at F, 
and for this reason the product of HB, BC together with the 
square on FC is equal to the square on BE. [11. 6] 

But, by hypothesis, the product of AB, BC together with 
the square on CE is also equal to the square on BE. 

Thus the product of HB, BC together with the square 
on CE will also be equal to the product of AB, BC together 
with the square on CE : 

which is absurd. 

Therefore the product of AB, BC together with the square 
on CE is not less than the square on BE. 

And it was proved that neither is it equal to the square 
on BE. 

Therefore the product of AB, BC together with the square 
on CE is not square. 

Q. E. D. 



We can, of course, write the identity in the note on Lemma 1 above (p. 64) 
in the simpler form 

where, as before, mp* t mf are both odd or both even. 
Now, says Euclid, 

mp* . mq* + vS— — *- - 1 J is not a square number. 

This is proved by rtductio ad absurdum. 



66 BOOK X TLemma 2, x. 29 

The number is dearly less than to/ 1 , mtf + (-^- — 1, i.e. less than 

If then the number is square, its side must be greater than, equal to, or 

fmt? + rap 1 V j. , ... , . w/ 1 + to/ 

less than I >* — —*• - 1 L the number next less than — — . 

But (1) the side can no l oe > I — — — 1) without being equal to 

-^- — —"- since they are consecutive numbers. 
2 

( 2 ) («/>- 2) TO?» + ^-~-— ~ l) = (-^7-^ " ') • I"- 6 1 

If then rnf.tnf+ C ~ M *~ - 1 J is also equal to f ffi*'"* _ , J j 

we must have {»»/* - 2) «/ = to/ 1 . to? 1 , 

or to/* - 2 = m/* : 

which is impossible. 

( 3) ir ^.^ + (-«!z*t , _ I ) , <(at±itf_ I ) , j 

suppose it equal to f - 1 - — r) . 

But [,,. 61 (to/ - «,) to/ + (5^tf _ r )' = f^W . r )* 
Therefore 

which is impossible. 

Hence all three hypotheses are false, and the sum of the squares 

j ■ j / m ?* ~ <*F \* ■ 
«(/• . toj* and I -i— — J- _ 1 1 is not square. 

Proposition 29, 

To find two rational straight lines commensurable in square 
only and such that the square on the greater is greater than 
the square on the less by the square on a straight line commen- 
surable in length with the greater. 

For let there be set out any rational straight line AB, 
and two square numbers CD, DE such that their difference 
CE is not square ; [Lemma 1] 

let there be described on AB the semicircle AFB, 




x. a 9 ] PROPOSITION 39 67 

and let it be contrived that, 

as DC is to CE, so is the square on BA to the square 
on AF. [x. 6, Pot.] 

Let FB be joined. 

Since, as the square on BA is to 
the square on AF, so is DC to CE, 
therefore the square on BA has to 
the square on AF the ratio which the 

number DC has to the number CE ; q g = 

therefore the square on BA is com- 
mensurable with the square on AF. [x. 6] 

But the square on AB is rational ; [x, Def. 4] 

therefore the square on AF is also rational ; [M.] 

therefore AFis also rational. 

And, since DC has not to CE the ratio which a square 
number has to a square number, 

neither has the square on BA to the square on AF the ratio 
which a square number has to a square number ; 
therefore AB is incommensurable in length with AF. [x. 9] 

Therefore BA, AF are rational straight lines commen- 
surable in square only. 

And since, as DC is to CE, so is the square on BA to 
the square on AF, 

therefore, converlendo, as CD is to DE, so is the square on 
AB to the square on BF. [v. 19, Por., m. 31, 1. 47] 

But CD has to DE the ratio which a square number has 
to a square number : 

therefore also the square on AB has to the square on BF 
the ratio which a square number has to a square number ; 
therefore AB is commensurable in length with BF. [x. 9] 

And the square on AB is equal to the squares on AF, FB ; 
therefore the square on AB is greater than the square on AF 
by the square on BF commensurable with AB. 

Therefore there have been found two rational straight 
lines BA, AF commensurable in square only and such that 
the square on the greater AB is greater than the square on 
the less AF by the square on BF commensurable in length 
with AB. 



68 BOOK X [x. 29, 30 

Take a rational straight line p and two numbers m\ «* such that (m* - n') 
is not a square. 

Take a straight line x such that 

»* , :iW , ,-« , = p , :* , (1), 

whence x? = r— p 1 , 



x=pifl^¥ t where i - 



Then p, p>Ji —& are the straight lines required. 

It follows from (1) that ^ " p*, 

and ;r is rational, but x <j p. 

By (1), cmverUndo, m* • «* = p* : p' - xt, 
so that tjp' -a? « p, and in fact = £p. 

According as p is of the form a or ^, the straight Hnes are (1) a, J a 1 - P 
or (a) JA, -J A - PA. 

Proposition 30. 

To find two rational straight lines commensurable in square 
only and such that the square on the greater is greater than 
the square on the less by the square on a straight line incom- 
mensurable in length with the greater. 

Let there be set out a rational straight line AB, 
and two square numbers CE, ED 
such that their sum CD is not 
square ; [Lemma a] 

let there be described on AB the 
semicircle AFB, 
let it be contrived that, 
as DC is to CE, so is the square 

on BA to the square on AF, <j e — — b 

[x. 6, Pot.] 
and let FB be joined. 

Then, in a similar manner to the preceding, we can prove 
that BA, AF are rational straight lines commensurable in 
square only. 

And since, as DC is to CE, so is the square on BA to 
the square on AF, 

therefore, convertendo, as CD is to DE, so is the square on 
AB to the square on BF. [v. 19, Por., in. 31, 1. 47] 

But CD has not to DE the ratio which a square number 
has to a square number ; 




x. 3°, 3'] PROPOSITIONS 29— 31 69 

therefore neither has the square on AB to the square on BF 
the ratio which a square number has to a square number ; 
therefore AB is incommensurable in length with BF. [x. 9] 

And the square on AB is greater than the square on AF 
by the square on FB incommensurable with A B, 

Therefore AB, AF are rational straight lines commen- 
surable in square only, and the square on AB is greater than 
the square on AF by the square on FB incommensurable in 
length with AB. 

q. e. a 

In this case we take m\ r? such that «' + »* is not square. 
Find x such that m* + «* : m 1 - p' : x 3 , 

—A 

whence x* = 



m" + n> ' 



Then />, 



x- p 


where k — — 
m 


itisfy the condition. 





•JWp 

The proof is after the manner of the proof of the preceding proposition 
and need not be repeated. 

According a s p is of the form a or J A, the straight lines take the 

y" %> a i~ 

a a ^ , that is, a, vV — B, or {2) J A, 'J A - B and 

JA, J~4^¥. 

Proposition 31. 

To find two medial straight lines commensurable in square 
only, containing a rational rectangle, and such that the square 
on the greater is greater than the square on the less by the 
square on a straight line commensurable in length with the 
greater. 

Let there be set out two rational straight lines A, B 
commensurable in square only and such that the 
square on A, being the greater, is greater than 
the square on B the less by the square on a 
straight line commensurable in length with A. 

[x. 29] 

And let the square on C be equal to the 
rectangle A, B. D 

Now the rectangle A, B is medial ; [x. 21] 
therefore the square on C is also medial ; 
therefore C is also medial. [x. ii] 



7 o BOOK X [x. 31 

Let the rectangle C, D be equal to the square on B. 

Now the square on B is rational ; 
therefore the rectangle C, D is also rational. 

And since, as A is to B, so is the rectangle A, B to the 
square on B, 

while the square on C is equal to the rectangle A, B, 
and the rectangle C, D is equal to the square on B, 
therefore, as A is to B, so is the square on C to the rectangle 
C,D. 

But, as the square on C is to the rectangle C, D, so is C 
toZ>; 
therefore also, as A is to B, so is C to D. 

But A is commensurable with B in square only ; 
therefore C is also commensurable with D in square only. [x. 1 1] 

And C is medial ; 
therefore D is also medial. [x. 23, addition] 

And since, as A is to B, so is C to D, 
and the square on A is greater than the square on B by the 
square on a straight line commensurable with A y 
therefore also the square on C is greater than the square on 
D by the square on a straight line commensurable with C. 

[x. 14] 

Therefore two medial straight lines C, D, commensurable 
in square only and containing a rational rectangle, have been 
found, and the square on C is greater than the square on D 
by the square on a straight Tine commensurable in length 
with C. 

Similarly also it can be proved that the square on C 
exceeds the square on D by the square on a straight line 
incommensurable with C, when the square on A is greater 
than the square on B by the square on a straight line incom- 
mensurable with A. [x. 30) 

I. Take the rational straight lines commensurable in square only found 
in x. 29, i.e. p, p Vi - ^. 

Take the mean proportional p(i - ¥)* and jc such that 
P (1 - #)* : p vT^P = p vT^P ! * 

Then P { 1 1 - #■)*, x, or p (i - &)*, p (i - 4?)* are straight lines satisfying the 
given conditions. 



x. 3i, 3*] PROPOSITIONS 31, 3a 71 

For (a) p'v'i -<P is a medial area, and therefore p(i -.**)* is a media! 

straight line (1); 

and x. p(i - #*)* = p*(i~&) and is therefore a rational area. 

(fi) p, p(i - J?)*, p V 1 - &, x are straight lines in continued proportion, by 
construction. 

Therefore p : p Vi -£* = p(r -#)* : x (a). 

(This Euclid has to prove in a somewhat roundabout way by means of the 
lemma after X. 21 to the effect that a : b = ab : P.) 

From (2) it follows [x. n] that x . — p(i -£*)*; whence, since p{\ -.4*)* is 

medial, x or p{i — £")* is medial also. 

(■y) From (2), since p, pVt -i* satisfy the remaining condition of the 

problem, p(i -£*)*, p{t-ff)° do so also [x. 14]. 

According as p is of the form a or J A, the straight lines take the forms 



(a) UA {A - PA), 



A-&A 



UA {A - PA) ' 

II. To find medial straight lines commens irable in square only contain- 
ing a rational rectangle, and such that the square on one exceeds the square 
on the other by the square on a straight line incommensurable with the former, 
we simply begin with the rational straight lines having the corresponding 

property [x. 30], viz. p, ■ ■ ■ , and we arrive at the straight lines 

VI +*F 

P P 



According as p is of the form a or ^/A, these {if we use the same 
transformation as at the end of the note on x. 30) may take any of the forms 

JTjj^, - a '~ jg _ . 



(1) 

(a) KlAlA^B), 

or \!A(A-P), 



A- B 

Va(A-b)' 

A-P 

Ma {A-p) 



Proposition 32. 

To find two medial straight lines commensurable in square 
only, containing a medial rectangle, and such that the square 
on the greater is greater than the square on the less Sy the 
square on a straight line commensurable with the greater. 



73 BOOK X fx. 31 

Let there be set out three rational straight lines A, B, C 
commensurable in square only, and such that the square on A 
is greater than the square on C by the square on a straight 
line commensurable with A, [x. 29] 

and let the square on D be equal to the rectangle A, B. 



Therefore the square on D is medial ; 
therefore D is also medial. [*■ "] 

Let the rectangle D, E be equal to the rectangle B, C. 

Then since, as the rectangle A, B is to the rectangle B, C, 
so is A to O, 

while the square on D is equal to the rectangle A, B, 
and the rectangle D, E is equal to the rectangle B, C, 
therefore, as A is to C, so is the square on D to the rectangle 
D,E. 

But, as the square on D is to the rectangle D, E, so is D 
to E ; 
therefore also, as A is to C, so is D to E. 

But A is commensurable with C in square only ; 
therefore D is also commensurable with E in square only. jx. 1 1] 

But D is medial ; 
therefore E is also medial. [x. 23, addition] 

And, since, as A is to C, so is D to E, 
while the square on A is greater than the square on C by 
the square on a straight line commensurable with A, 
therefore also the square on D will be greater than the square 
on E by the square on a straight line commensurable with D. 

[x. 14] 

I say next that the rectangle D, E is also medial. 

For, since the rectangle B, C is equal to the rectangle D, E, 
while the rectangle B, C is medial, [x. 31] 

therefore the rectangle D, E is also medial. 

Therefore two medial straight lines D, E, commensurable 
in square only, and containing a medial rectangle, have been 
found such that the square on the greater is greater than the 



x. 3*] PROPOSITION 32 73 

square on the less by the square on a straight line commen- 
surable with the greater. 

Similarly again it can be proved that the square on D 
is greater than the square on E by the square on a straight 
line incommensurable with D, when the square on A is 
greater than the square on C by the square on a straight line 
incommensurable with A. [x. 30] 



I. Euclid takes three straight lines of the form p, p Jk, p vi-/*, 

takes the mean proportional pk* between the first two ( i ), 

and then finds -t such that 

pX i :p^ = p^i^¥:x (2), 

whence x = pk* Ji -M, 

and the straight lines pk*, pk* Ji — 4? satisfy the given conditions. 
Now (a) pk* is medial. 
(#) We have, from (i) and {2), 

p:p*Ji-# = pk i :x (3), 

whence x . — pk* ; and x is therefore medial and «- pk*. 

(y) x.pk* = Pt Jk.pJi-£'. 

But the latter is media! ; [x. 2 1 J 

therefore x . pk*, or pk* . pk* Ji - JP, is medial. 

Lastly (8) p, p Ji — ¥ have the remaining property in the enunciation ; 
therefore pk*, pX <Ji —JP have it also. [x. 14] 

(Euclid has not the assistance of symbols to prove the proportion {3) above. 
He therefore uses the lemmas ab;bc=a\c and d 1 \de = d\e to deduce from 
the relations 

06 = 4* \ 
and d : b = e : e J 

that a:c = d:e.) 

The straight lines pk*, p\*Ji-& may take any of the following forms 
according as the straight lines first taken are 

(1) «i JB, j&±&, f» JA, JB, JA^VA , (3) JA, b, J A -PA. 

, jB{a>-S) 

(1) JaJB, 

(2) $AB, 

(3) -Jb~[A, 



•JaJB ' 
>JB(A-#A) 
ifAB ' 




bJA-#A 

•JbJA 





74 



BOOK X 



[x. 3*, Lemma 



II. If the other conditions are the same, but the square on the first 
medial straight line is to exceed the square on the second by the square on a 
straight line ituommensxrable with the first, we begin with the three straight 

lines p, pj\, -,—-- , and the medial straight lines are 
vi +# 

->* 

P ki, 



The possible forms are even more various in this case owing to the more 
various forms that the original lines may take, e.g. 

(I) a, JB> Jtf^Ci 

(z) J A, b, -JA~-?; 

(3) sM> 4 -Ja^C; 

< 4 ) J A, JB, JA-?\ 
($) JA, JB, JJ^C; 
the medial straight lines corresponding to these being 

slB(a*-C). 



(r) J^JB, 

(2) JFJA, 

(3) VTO 

( 4 ) Has, 

(s) $2B, 



■JTjB ' 

b Ja-c' 
JbijA ' 

b-J A~C 
Jb~J2 ' 

JB{A~^) 

Hab ' 
JbJaJ^C) 
Sab ' ' 



Lemma. 

Let ABC be a right-angled triangle having the angle A 
right, and let the perpendicular AD be 
drawn ; A 

I say that the rectangle CB, BD is 
equal to the square on BA, 
the rectangle BC, CD equal to the 
square on CA, 

the rectangle BD, DC equal to the square on AD, 

and, further, the rectangle BC, AD equal to the rectangle 




Lemma, x. 33] PROPOSITIONS 33, 33 75 

And first that the rectangle CB, BD is equal to the square 
on BA. 

For, since in a right-angled triangle AD has been drawn 
from the right angle perpendicular to the base, 
therefore the triangles ABD, ADC are similar both to the 
whole ABC and to one another. [vi. 8] 

And since the triangle ABC is similar to the triangle ABD, 
therefore, as CB is to BA, so is BA to BD ; [vi. 4] 

therefore the rectangle CB, BD is equal to the square on AB. 

[vi. 17] 

For the same reason the rectangle BC, CD is also equal 
to the square on AC. 

And since, if in a right-angled triangle a perpendicular 
be drawn from the right angle to the base, the perpendicular 
so drawn is a mean proportional between the segments of the 
base, [vi. 8, Por.] 

therefore, as BD is to DA, so is AD to DC; 
therefore the rectangle BD, DC is equal to the square on AD. 

[vi. 17] 

I say that the rectangle BC, AD is also equal to the rect- 
angle BA, AC. 

For since, as we said, ABC is similar to ABD, 
therefore, as BC is to CA, so is BA to AD. [vi. 4] 

Therefore the rectangle BC, AD is equal to the rectangle 
BA, AC. [vi. 16] 

Q. E. D. 

Proposition 33. 

To find two straight lines incommensurable in square which 
make the sum of the squares on them rational hut the rectangle 
contained by them medial. 

Let there be set out two rational straight lines AB, BC 
commensurable in square only 
and such that the square on the 
greater A Bis greater than the 
square on the less BC by the 
square on a straight line in- 
commensurable with AB, 

[*• 3°] 




76 BOOK X [x. 33 

let BC be bisected at D, 

let there be applied to AB a parallelogram equal to the square 

on either of the straight lines BD, DC and deficient by a 

square figure, and let it be the rectangle AE, EB j [vi. 18] 

let the semicircle AFB be described on AB, 

let EF be drawn at right angles to AB, 

*nd let AF, FB be joined. 

Then, since AB, BC are unequal straight lines, 

and the square on AB is greater than the square on BC by 
the square on a straight line incommensurable with AB, 

while there has been applied to AB a parallelogram equal to 
the fourth part of the square on BC, that is, to the square on 
half of it, and deficient by a square figure, making the rect- 
angle AE, EB, 
therefore AE is incommensurable with EB. [x. 18] 

And, as AE is to EB, so is the rectangle BA, AE to the 
rectangle AB, BE, 

while the rectangle BA, AE is equal to the square on AF, 
and the rectangle AB, BE to the square on BF; 

therefore the square on AF is incommensurable with the 
square on FB ; 

therefore AF, FB are incommensurable in square. 

And, since AB is rational, 
therefore the square on AB is also rational ; 
so that the sum of the squares on AF, FB is also rational. 

[>■ 47] 

And since, again, the rectangle AE, EB is equal to the 
square on EF, 

and, by hypothesis, the rectangle AE, EB is also equal to the 
square on BD, 

therefore FE is equal to BD : 

therefore BC is double of FE, 

so that the rectangle AB, BC is also commensurable with the 
rectangle AB, EF. 

But the rectangle AB, BC is medial ; [x. n] 

therefore the rectangle AB, EF is also medial. [x. 23, Por.] 



x. 33) PROPOSITION 3 3 77 

But the rectangle AB, EF is equal to the rectangle AF, 

FB ', [Lemma] 

therefore the rectangle AF, FB is also medial. 

But it was also proved that the sum of the squares on these 
straight lines is rational. 

Therefore two straight lines AF, FB incommensurable 
in square have been found which make the sum of the 
squares on them rational, but the rectangle contained by them 
medial. 

Q. E. D, 

Euclid take* the straight lines found in x. 30, viz. p, 



*/! + #' 

He then solves geometrically the equations 

x+y = p 



xy = - 



■(•)• 



If x, y are the values found, he takes u, v such that 

*:%} ■■ » 

and u, v art; straight lines satisfying the conditions of the problem. 
Solving algebraically, we get (if x >_y) 



whence 



^y ; 



V* v s/TT* 1 



"AV^ 



•(3)- 



V 2 v Vi+#, 

Euclid's proof that these straight lines fulfil the requirements is as follows. 

(a) The constants in the equations (1) satisfy the conditions of x. 18 ; 
therefore x ^y. 

But x :y = u* :tf. 

Therefore «* « e*, 

and a, v are thus incommensurable in square. 

(0) ** + = p*, which is rational. 
( r ) By (1), V^= / ■ 

By (1), uv = p.Jxy 






78 BOOK X [x. 33, 34 

But — ?£=. is a medial area. 

therefore uv is medial. 

Since p, ■ .: - may have any of the three forms 

(!) a, JfZTB, K ,) JA, Ja^B, (3) JA, -JA^P, 
u, v may have any of the forms 

, , teF+ajM Is'-ajB 

M V --5— ■ V-* ' 

/a + >Ja£ I a- 4ab 

<«) V ~ > V - - ■ ; 

, . fATFJl /a~VJa 

w V — , V — - — • 

Proposition 34. 

To find two straight lines incommensurable in square which 
make the sunt of the squares on them medial but the rectangle 
contained by them rational. 

Let there be set out two medial straight lines AB, BC, 
commensurable in square only, such that the rectangle which 
they contain is rational, and the square on AB is greater than 
the square on BC by the square on a straight line incom- 
mensurable with AB ; [x. 31, adfin.\ 




let the semicircle ADB be described on AB, 

let BC be bisected at E, 

let there be applied to AB a parallelogram equal to the square 

on BE and deficient by a square figure, namely the rectangle 

AF, FB; [vi. 28] 

therefore AF is incommensurable in length with FB. [x. 18] 

Let FD be drawn from F at right angles to AB, 
and let AD, DB be joined. 



x. 34] PROPOSITIONS 33. 34 79 

Since AF is incommensurable in length with FB, 

therefore the rectangle BA, AF is also incommensurable with 
the rectangle AB, BF. [x. 1 1] 

But the rectangle BA, AF \s equal to the square on AD, 
and the rectangle AB, BF to the square on DB ; 
therefore the square on AD is also incommensurable with the 
square on DB. 

And, since the square on AB is medial, 
therefore the sum of the squares on AD, DB is also medial. 

[in. 31, 1. 47] 

And, since BC is double of DF, 

therefore the rectangle AB, BC is also double of the rectangle 
AB, FD. 

But the rectangle AB, BC is rational ; 
therefore the rectangle AB, FD is also rational. [x. 6] 

But the rectangle AB, FD is equal to the rectangle AD, 
DB ; [Lemma] 

so that the rectangle AD, DB is also rational. 

Therefore two straight lines AD, DB incommensurable 
in square have been found which make the sum of the squares 
on them medial, but the rectangle contained by them rational, 

Q E. D. 
In this case we take [x. 31, and part] the medial straight lines 
P P 

Solve the equations 

P \ 

x+y = , 

(1 +&)* 



Take u, v such that, if x, y be the result of the solution, 



.(!>. 



and u, v are straight lines satisfying the given conditions. 

Euclid's proof is similar to the preceding, 
(a) From {() it follows [x. 18] that 

x„y, 
whence u* v »*, 

and <r, v are thus incommensurable in square. 



■(*)■ 



So BOOK X [x. 34, 35 

(Jl) a* + 1^ — -£ — which is a medial area. 

IP 1 . 

= - . — - .. . which is a rational area. 

Therefore uv is rational. 

To find the actual form of u, v, we have, by solving the equations (i) 
(if x>y), 

2 — iiJT+P + i), 



x = 






and hence « = -■ p - -■■ •JjT^W+k. 



Ja(iTF) 



Bearing in mind the forms which — £ — - , - — * may take (see note 

ft + iPJ* (i+#)* 
on x, 31), we shall find that u, v may have any of the forms 

, /{ a + JB) «/^~g J ( a -JB)>]7=li _ 



, . / uA + JB)*lA-£ hjA-JB)JA-B . 

Proposition 35. 

To find two straight lines incommensurable in square which 
make the sum of the squares on them medial and the rectangle 
contained by them medial and moreover incommensurable with 
the sum of the squares on them. 

Let there be set out two medial straight lines AB, BC 
commensurable in square only, containing a medial rectangle, 
and such that the square on AB is greater than the square on 
BC by the square on a straight line incommensurable with 
AB; [x.$t,a£Jkt.] 



x. 35] PROPOSITIONS 34, 35 81 

let the semicircle ADB be described on AB, 
and let the rest of the construction be as above. 




Then, since AF is incommensurable in length with FB, 

[x. 18] 
AD is also incommensurable in square with DB. [x. u] 

And, since the square on AB is medial, 
therefore the sum of the squares on AD, DB is also medial. 

[in. 31, 1.47] 
And, since the rectangle AF, FB is equal to the square 
on each of the straight lines BE, DF, 

therefore BE is equal to DF; 

therefore BC is double of FD, 

so that the rectangle AB, BC is also double of the rectangle 
AB, FD. 

But the rectangle AB, BC is medial ; 

therefore the rectangle AB, FD is also medial. [x, 3*, Por.] 

And it is equal to the rectangle AD, DB ; 

[Lemma after x. 32] 

therefore the rectangle AD, DB is also medial. 

And, since AB is incommensurable in length with BC, 
while CB is commensurable with BE, 
therefore AB is also incommensurable in length with BE, 

[x. 13] 
so that the square on AB is also incommensurable with the 
rectangle AB, BE. [x. n] 

But the squares on AD, DB are equal to the square on 

AB, [K47] 

and the rectangle AB, FD, that is, the rectangle AD, DB, is 
equal to the rectangle AB, BE; 

therefore the sum of the squares on AD, DB is incommen- 
surable with the rectangle AD, DB. 



Ra BOOK X [x. 35 

Therefore two straight lines AD, DB incommensurable 
in square have been found which make the sum of the squares 
on them medial and the rectangle contained by them medial 
and moreover incommensurable with the sum of the squares 
on them. 

Q, E. D. 

Take the medial straight lines found in x. 3a {and part), viz. 

pA*, p\ i /^/i^ i . 
Solve the equations 

x+y- pX* 

„ _p\A [ (I> ' 

and then put u*- r . 



« , = pX i .jcl 

ll> = pk* .y \ 



where x, y are the ascertained values of x, y. 

Then u, v. are straight lines satisfying the given conditions. 
Euclid proves this as follows. 

(«) From (1) it follows [x. 18] that x ^y. 
Therefore k* ^ t^, 

arid u ~- v. 

(p3) v* + »" = p , ,/A, which is a medial area (3). 

(y) uv = p\* . <Jxy 

I n* /X 

= - ■ / -^— ■ ) which is a medial area (4); 

therefore uv is medial. 

(«) t**~i-dL, 

whence ?Jk~~ -4^* ■ 

That is, by (3) and (4}, 

The actual values are found thus. Solving the equations (1), we have 
pA*/ i \ 

pX* / h \ 

whence ,.!*/, 4^ , 

pA* / A 



x. 35. 3«] PROPOSITIONS 35, 3° 83 

According as p is of the form a or J A, we have a variety of forms for 
u, v, arrived at by using the same transformations as in the notes on X. 30 
and X. 3* {second part), e.g. 

( 3 ) J SBH Ji, /QaEam. 

and the expressions in (z), (3) with b in place of JB. 

Proposition 36. 
If two rational straight lines commensurable in square 

only be added together, the whole is irrational; and let it be 
called binomial. 

For let two rational straight lines AB, BC commen- 
5 surable in square only be added 
together ; 

I say that the whole AC is ir- A 8 c 

rational. 

For, since AB is incommensurable in length with BC — 
10 for they are commensurable in square only — 
and, as AB is to BC, so is the rectangle AB, BC to the 
square on BC, 

therefore the rectangle AB, BC is incommensurable with the 

square on BC. [x. itj 

15 But twice the rectangle AB, BC is commensurable with 

the rectangle AB, BC [x. 6], and the squares on AB, BC are 

co- n mensurable with the square on BC— for AB, BC are 

rational straight lines commensurable in square only — [x. 15] 

therefore twice the rectangle AB, BC is incommensurable 

» with the squares on AB, BC. [x. 13] 

And, componendo, twice the rectangle AB, BC together 

with the squares on AB, BC, that is, the square on AC [n. 4], 

is incommensurable with the sum of the squares on AB, BC. 

[x .6] 
But the sum of the squares on AB, BC is rational ; 

25 therefore the square on A C is irrational, 
so that AC is also irrational. [x. Def. 4] 

And let it be called binomial. 



84 BOOK X [x.36, 37 

Here begins the first hexad of propositions relating to compound irrational 
straight lines. The six compound irrational straight lines are formed by 
adding two parts, as the corresponding six in Props. 73 — 78 are formed by 
subtraction, 1'he relation between the six irrational straight lines in this and 
the next five propositions with those described in Definitions 11. and the 
Props. 48 — S3 following thereon (the first, second, third, fourth, fifth and 
sixth binomials) will be seen when we come to Props. 54 — 59 ; but it may be 
stated here that the six compound irrationals in Props. 36 — 41 can be found 
by means of the equivalent of extracting the square root of the compound 
irrationals in x. 48 — 53 (the process being, strictly speaking, the finding of the 
sides of the squares equal to the rectangles contained by the latter irrationals 
respectively and a rational straight line as the other side), and it is therefore 
the further removed compound irrational, so to speak, which is treated first. 

In reproducing the proofs of the propositions, I shall for the sake of 
simplicity call the two parts of the compound irrational straight line x, y, 
explaining at the outset the forms which x, y really have in each case 5 x will 
always be supposed to be the greater segment. 

In this proposition x, y are of the form p, Jk . p, and (x +y) is proved to 
be irrational thus. 

x ~-y, so that x uj. 

Now x :y = x t : xy, 

so that x* v xy. 

But x 1 f (x* +y), and xy « ixy ; 
therefore (.x* +_>■*) « 2xy, 

and hence («" +y + axy) « (a^ +y). 

But (x? +y) is rational ; 
therefore (x +yf, and therefore (* +y), is irrational. 

This irrational straight line, p + Jk . p, is called a binomial straight line. 

This and the corresponding apatomt (p — ^/Jb . p) found in x. 73 are the 
positive roots of the equation 

Proposition 37. 

ff two medial straight lines commensurable in square only 
and containing a rational rectangle be added together, the 
whole is irrational ; and let it be called a first bi medial 
straight line. 

For let two medial straight lines AB, BC commensurable 
in square only and containing 

a rational rectangle be added ^ g — — — c 

together ; 

] say that the whole AC is irrational. 

For, since AB is incommensurable in length with BC, 
therefore the squares on AB, BC are also incommensurable 
with twice the rectangle AB, BC; [cf. x. 36, 11. 9— *o] 



x 37- 3»] PROPOSITIONS 36—38 85 

and, componendo, the squares on AB, BC together with twice 
the rectangle AB, BC, that is, the square on AC [11. 4], is 
incommensurable with the rectangle AB, BC. [x. 16] 

But the rectangle AB, BC is rational, for, by hypothesis, 
AB, BC are straight lines containing a rational rectangle ; 
therefore the square on AC is irrational ; 
therefore AC is irrational. [x. Def. 4] 

And let it be called a first bimedial straight line. 

Q. E. D, 

Here x,y have the forms &*p, $p respectively, as found in x. 27. 

Exactly as in the last case we prove that 
x* +_y* u zxy, 
whence (* +y)* « txy. 

But xy is rational , 
therefore (x +yf, and consequently (x +y), is irrational. 

The irrational straight line trp + k*p is called a first bimedial straight line. 

This and the corresponding first apotome of a medial (Jrp - A*p) found in 
x. 74 are the positive roots of the equation 

X* - 2 jk < I + k) p* . x' + k (1 - kfp',= o. 



Proposition 38. 

If two medial straight lines commensurable in square only 
and containing a medial rectangle be added together, the whole 
is irrational; and let it be called a second bimedial straight 
line. 

5 For let two medial straight lines AB, BC commensurable 

in square only and containing 

a medial rectangle be added a B o 

together; p H Q 

I say that AC is irrational. 
10 For let a rational straight 

line DE be set out, and let the 

parallelogram DF equal to the 

square on AC be applied to DE, 

producing DG as breadth. [1. 44] 

15 Then, since the square on A C is equal to the squares on 

AB, BC and twice the rectangle AB, BC, [11. 4] 

let EH, equal to the squares on AB, BC, be applied to DE; 







E 


F 



86 BOOK X [x. 38 

therefore the remainder HF is equal to twice the rectangle 
AB, BC. 

jo And, since each of the straight lines AB, BC is medial, 
therefore the squares on AB, BC are also medial. 

But, by hypothesis, twice the rectangle AB, BC is also 
medial. 

And EH is equal to the squares on AB, BC, 

»s while FH is equal to twice the rectangle AB, BC; 

therefore each of the rectangles EH, HF is medial. 

And they are applied to the rational straight line DE ; 

therefore each of the straight lines DH, HG is rational and 
incommensurable in length with DE. [x. n] 

3f> Since then AB is incommensurable in length with BC, 

and, as AB is to BC, so is the square on AB to the rectangle 
AB, BC, 

therefore the square on AB is incommensurable with the rect- 
angle AB, BC. [x. 11] 

35 But the sum of the squares on AB, BC is commensurable 
with the square on AB, [x. 15] 

and twice the rectangle AB, BC Is commensurable with the 
rectangle AB, BC. [x. 6] 

Therefore the sum of the squares on AB, BC is incom- 
40 mensurable with twice the rectangle AB, BC. [x. 13] 

But EH is equal to the squares on AB, BC, 

and HF is equal to twice the rectangle AB, BC. 

Therefore EH is incommensurable with HF, 

so that DH is also incommensurable in length with HG. 

[vi. t, x. 1 1] 

45 Therefore DH, HG are rational straight lines commen- 
surable in square only ; 

so that DG is irrational. [x. 36] 

But DE is rational ; 

and the rectangle contained by an irrational and a rational 
50 straight line is irrational ; [cf. x. 20] 

therefore the area DF is irrational, 
and the side of the square equal to it is irrational. [x. Def. 4] 



x. 3 8, 39] PROPOSITIONS 38, 39 87 

But AC is the side of the square equal to DF\ 
therefore AC is irrational. 
s$ And let it be called a second bimedial straight Hue. 

Q. E. D. 

After proving (1. 2 1 ) that each of the squares on AB, BC is medial, Euclid 
states (II. 24, 26) that EH, which is equal to the sum of the squares, is a 
medial area, but does not explain why. It is because, by hypothesis, the 
squares on AB, BC are commensurable, so that the sum of the squares is 
commensurable with either [x. 15] and is therefore a medial area [x. 23, Por.J. 

In this case [x. 28, note] x, y are of the forms Irp, X'p/^ respectively. 
Apply each of the areas (x* +y) and zxy to a rational straight line <r, i.e. 
suppose 

axy = av. 

Now it follows from the hypothesis, x. 15 and x. 13, Por. that (x t +y t ) is 
a medial area ; and so is 2 xy, by hypothesis ; 
therefore vu, av are medial areas. 

Therefore each of the straight lines u, v is rational and w a (r). 

Again x u y ; 

therefore x* u xy. 

But x* " x* +y* and xy « txy ; 
therefore x'+y i v3xy, 
or <ru ^ av, 
whence « ^ v (*)■ 

Therefore, by (t), (2), u, ware rational and «-. 

It follows, by x. 36, that {« + v) is irrational. 

Therefore {u + v) a is an irrational area [this can be deduced from x, ao 
by reductio ad absurdum\ 

whence (x +y) 1 , and consequently (x+y), is irrational. 

The irrational straight line J^p + — j is called a second bimedial straight 
tine. 

This and the corresponding second apotome of a media! (irp— ^r p) 
found in x. 75 are the positive roots of the equation 



^-^^A?-** 



p' 



Proposition 39. 

If two straight lines incommensurable in square which 
make the sum of the squares on them rational, but the rectangle 
contained by them medial, be added together, the whole straight 
line is irrational : and let it be called major. 



88 BOOK X [x. 39, 40 

For let two straight lines AB, BC incommensurable in 
square, and fulfilling the given con- 
ditions [x. 33], be added together; j g 6 
I say that AC is irrational. 

For, since the rectangle AB, BC is medial, 
twice the rectangle AB, BC is also medial. [x. 6 and 23, Por.] 

But the sum of the squares on AB, BC is rational ; 
therefore twice the rectangle AB, BC is incommensurable 
with the sum of the squares on AB, BC, 
so that the squares on AB, BC together with twice the rect- 
angle AB, BC, that is, the square on AC, is also incommen- 
surable with the sum of the squares on AB, BC ; [x. 16] 

therefore the square on A C is irrational, 
so that AC is also irrational, [x Def. 4] 

And let it be called major. 

Q. E. D. 
Here x, y are of the form found in x. 33, vnz. 

j*v 1 + jt+p' V2V 1 vn^r 

By hypothesis, the rectangle xy is media] , 
therefore txy is medial. 

Also [x* +y T ) is a rational area. 

Therefore x* -i-y* « ixy, 

whence (x +_y) s « (x* +y% 

so that (x +y)\ and therefore (x +y), is irrational. 

The irrational straight line -?- */ 1 

called a major (irrational) straight line. 

This and the corresponding minor irrational found in x. 76 are the 

positive roots of the equation 

W 

x*—2p'.x' + ii^a 

r 1 + P r 



Proposition 40. 

If two straight lines incommensurable in square which 
make the sum of the squares on them medial, but the rectangle 
contained by them rational, be added together, the whole straight 
line is irrational ; and let it be called the side of a rational 
plus a medial area. 



X. 40,41] PROPOSITIONS 39— 41 »9 

For let two straight lines AB, BC incommensurable in 
square, and fulfilling the given con- 
ditions [x. 34], be added together ; a b o 

I say that AC is irrational. 

For, since the sum of the squares on AB, BC is medial, 
while twice the rectangle AB, BC is rational, 
therefore the sum of the squares on AB, BC is incommen- 
surable with twice the rectangle AB, BC; 
so that the square on AC is also incommensurable with twice 
the rectangle AB, BC, [x. 16] 

But twice the rectangle AB, BC is rational ; 
therefore the square on A C is irrational. 

Therefore AC is irrational. [x. Def. 4] 

And let it be called the side of a rational plus a 
medial area. 

Q. E. D. 
Here x,y have [x. 34] the forms 



In this case (x* +y) is a medial, and ixy a rational, area ; thus 

** +j? v ixy. 
Therefore (x +yf ^ axy, 

whence, since sxy is rational, 

(x + y) t , and consequently (x +^) t is irrational. 
The irrational straight line 

is called {for an obvious reason) the " side " of a rational plus a medial (area). 
This and the corresponding irrational with a minus sign found in x. 77 
are the positive roots of the equation 

x*- . . -- 2 p*.x*+ . ^,. K . p l = o. 

Proposition 41. 

If two straight lines incommensurable in square which 
make the sum 0/ the squares on them medial, and the rectangle 
contained by them medial and also incommensurable with the 
sum of the squares on them, be added together, Ike whole straight 
line is irrational; and let it be called the side of the sum 
of two medial areas. 



90 BOOK X [x. 41 

For let two straight lines AB, BC incommensurable in 
square and satisfying the given conditions 
[x. 35] be added together ; 
I say that AC is irrational. 

Let a rational straight line DE be set out, 
and let there be applied to DE the rectangle 
DF equal to the squares on A B, BC, and 
the rectangle GH equal to twice the rectangle 
AB,BC; 

therefore the whole DH is equal to the square 
on AC. [h- 4] 

Now, since the sum of the squares on 
AB, BC is medial, /" 

and is equal to DF, A S ~~° 

therefore DF is also medial. 

And it is applied to the rational straight line DE ; 
therefore DG is rational and incommensurable in length with 
DE. [x. «] 

For the same reason GK is also rational and incommen- 
surable in length with GF, that is, DE. 

And, since the squares on AB, BC are incommensurable 
with twice the rectangle AB, BC, 
DF is incommensurable with GH; 
so that DG is also incommensurable with GK. [vi. t, x. u] 

And they are rational ; 
therefore DG, GK are rational straight lines commensurable 
in square only ; 
therefore DK is irrational and what is called binomial, [x. 36] 

But DE is rational ; 
therefore DM is irrational, and the side of the square which 
is equal to it is irrational. [x. Def. 4] 

But AC is the side of the square equal to HD ; 
therefore AC is irrational. 

And let it be called the side of the sum of two medial 
areas. 



In this case x, y are of the form 

jW 1 + jtt§' j* v 



Q. E. D. 



4» V sfT+T** J* v JTT#' 



x. 4i, Lemma] PROPOSITION 41 91 



l=<M n (.). 



By hypothesis, (x 1 + ,r) and % xy are medial areas, and 

x* +y* v 2xy (1). 

' Apply ' these areas respectively to a rational straight line <r, and suppose 
x* +y — <ru 1 
axy ■■ 

Since then au and av are both medial areas, u, v are rational and both 
are ^ a (3). 

Now, by (1) and (i), 

so that guv. 

By this and (3), «, w are rational and "-. 
Therefore [x. 36] (» + v) is irrational. 
Hence <r(u +v) is irrational [deduction from x. 20]. 
Thus (jc +J>Y, and therefore (x + y), is irrational. 
The irrational straight line 



?iV V.+-4* %/" V 



is called (again for an obvious reason) the "side" of the sum 0/ two mtttials 
(medial areas). 

This and the corresponding irrational with a minus sign found in X. 78 
are the positive roots of the equation 

& 

X* - 2 Jk . X'p' + k r; p* = 0. 

I + K 

Lemma. 

And that the aforesaid irrational straight lines are divided 
only in one way into the straight lines of which they are the 
sum and which produce the types in question, we will now 
prove after premising the following lemma. 

Let the straight fine AB be set out, let the whole be cut 
into unequal parts at each of 

the points C, D, . 

and let^fTbe supposed greater A dec 

than DB ; 

I say that the squares on A C, CB are greater than the squares 
on AD, DB. 

For let AB be bisected at E. 

Then, since AC is greater than DB, 
let DC be subtracted from each ; 
therefore the remainder AD is greater than the remainder CB, 

But AE is equal to EB ; 
therefore DE is less than EC ; 



91 BOOK X [Lemma, x. 41 

therefore the points C, D are not equidistant from the point 
of bisection. 

And, since the rectangle AC, CB together with the square 
on EC is equal to the square on EB, [»■ 5] 

and, further, the rectangle AD, DB together with the square 
on DE is equal to the square on EB, [id.] 

therefore the rectangle AC, CB together with the square on 
EC is equal to the rectangle AD, DB together with the 
square on DE. 

And of these the square on DE is less than the square 
on EC; 

therefore the remainder, the rectangle AC, CB, is also less 
than the rectangle AD, DB, 

so that twice the rectangle AC, CB is also less than twice 
the rectangle AD, DB. 

Therefore also the remainder, the sum of the squares on 
AC, CB, is greater than the sum of the squares on AD, DB. 

Q. E, D. 

3. and which produce the types in question. The Greek is rnnnmar t& vpttflpcvti 
rfi^p and I have taken eKij to mean "types (of irrational straight lines)," though the expression 
might perhaps mean " satisfying the conditions in question," 

This proves that, if x +y = u + v, and if u, ti are more nearly equal than 
x, y {Le. if the straight line is divided in the second case nearer to the point 
of bisection), then 

(*•+/)>(«* + i/«). 
It is first proved by means of 11. 5 that 

zxy < tuv, 
whence, since (x +yy = (u + v)\ the required result follows. 



Proposition 42. 

A binomial straight line is divided into its terms at one 
point only. 

Let AB be a binomial straight line divided into its terms 
at C; 

therefore AC, CB are rational ^ -5 g b 

straight lines commensurable in 

square only. [*■ 3t>] 

I say that AB is not divided at another point into two 
rational straight lines commensurable in square only. 



x. 4 i] PROPOSITION 43 93 

For, if possible, let it be divided at D also, so that AD, 
DB are also rational straight lines commensurable in square 
only. 

It is then manifest that AC is not the same with DB. 

For, if possible, let it be so. 

Then AD will also be the same as CB, 

and, as AC is to CB, so will BD be to DA; 

thus AB will be divided at D also in the same way as by the 
division at C: 

which is contrary to the hypothesis. 

Therefore AC is not the same with DB. 

For this reason also the points C, D are not equidistant 
from the point of bisection. 

Therefore that by which the squares on AC, CB differ 
from the squares on AD, DB is also that by which twice 
the rectangle AD, DB differs from twice the rectangle 

AC, CB, 

because both the squares on AC, CB together with twice the 
rectangle AC, CB, and the squares on AD, DB together 
with twice the rectangle AD, DB, are equal to the square 
on AB. [a. 4] 

But the squares on AC, CB differ from the squares on 

AD, DB by a rational area, 

for both are ration, 1! ; 

therefore twice the rectangle AD, DB also differs from twiGe 
the rectangle AC, CB by a rational area, though they are 
medial [x. n] : 

which is absurd, for a medial area does not exceed a medial 
by a rational area. [x. 26] 

Therefore a binomial straight line is not divided at different 
points ; 
therefore it is divided at one point only. 

Q. E. D. 

This proposition proves the equivalent of the well-known theorem in 
surds that, 

if a + Ji - x + Jy, 

then a = x, b =y, 

and if Ja + Ji = Jx+- Jy, 

then a = x, b=y (or a=y, h = x). 



94 BOOK X [x. 42, 43 

The proposition states that a binomial straight line cannot be split up into 
terms (ovo/wto.) in two ways. For, if possible, let 

x+y-x' +y, 
where x, y, and also x', y\ are the terms of a binomial straight line, x', y' 
being different from x, y (or y, x). 

One pair is necessarily more nearly equal than the other. Let x',y be 
more nearly equal than x, y, 

Then (a* +y*) - (*' a +/*) = tx'y' - 2xy. 

Now by hypothesis (x? +y), (xP +y ) are rational areas, being of the form 

p'+V; 

but 2x'y, zxy are medial areas, being of the form ^jk.ff; 
therefore the difference of two medial areas is rational i 
which is impossible. [x. 26] 

Therefore x", y cannot be different from x, y (oi'y, x r 



Proposition 43. 

A first bimedial straight line is divided at one point only. 

Let AB be a first bimedial straight line divided at C, so 
that AC, CB are medial straight 

lines commensurable in square , — ■ . 

only and containing a rational A ° c B 

rectangle ; [x. 37] 

I say that AB is not so divided at another point. 

For, if possible, let it be divided at D also, so that AD, 
DB are also medial straight lines commensurable in square 
only and containing a rational rectangle. 

Since, then, that by which twice the rectangle AD, DB 
differs from twice the rectangle AC, CB is that by which the 
squares on AC, CB differ from the squares on AD, DB, 
while twice the rectangle AD, DB differs from twice the 
rectangle AC, CB by a rational area — for both are rational — 

therefore the squares on A C, CB also differ from the squares 
on AD, DB by a rational area, though they are medial : 

which is absurd, [x. 16] 

Therefore a first bimedial straight line is not divided into 
its terms at different points ; 

therefore it is so divided at one point only. 



43i 44] 



PROPOSITIONS 42—44 



95 






In this case, with the same hypothesis, viz. that 

x +_v = *' +y, 

and *', / are more nearly equal than x, y, 

we have as before (x* +y") - (x'*+y u ) = zx'y - 2xy. 

But, from the given properties of x, y, and at, y', it follows that txy, sx'y 
are rational, and (x* +>*), (x'' + / s ) medial, areas. 

Therefore the difference between two medial areas is rational : 
which is impossible. [x. 26] 



Proposition 44. 
A second bimedial straight line is divided at one point only. 

Let AB be a second bimedial straight line divided at C, 
so that AC, CB are medial straight lines commensurable in 
square only and containing a medial rectangle ; [x. 38] 

it is then manifest that C is not at the point of bisection, 
because the segments are not commensurable in length. 

I say that AB is not so divided at another point. 

A O C B 



C 

— 1— 



M 



F L Q K 



For, if possible, let it be divided at D also, so that AC is 
not the same with DB, but AC is supposed greater ; 
it is then clear that the squares on AD, DB are also, as we 
proved above [Lemma], less than the squares on AC, CB ; 
and suppose that AD, DB are medial straight lines commen- 
surable in square only and containing a medial rectangle. 

Now let a rational straight line EF be set out, 
let there be applied to EF the rectangular parallelogram EK 
equal to the square on AB, 

and let EG equal to the squares on AC, CB be subtracted ; 
therefore the remainder HK is equal to twice the rectangle 
AC, CB, [it. 4] 

Again, let there be subtracted EL, equal to the squares 
on AD, DB, which were proved less than the squares on 
AC, CB [Lemma] ; 



96 BOOK X [x. 44 

therefore the remainder MK is also equal to twice the rect- 
angle AD, DB. 

Now, since the squares on AC, CB are medial, 
therefore EG is medial. 

And it is applied to the rational straight line EF; 

therefore EH is rational and incommensurable in length with 
EF, [x, 22] 

For the same reason 

HN is also rational and incommensurable in length with EF. 

And, since AC, CB are medial straight lines commen- 
surable in square only, 
therefore AC vs. incommensurable in length with CB, 

But, as A C is to CB, so is the square on AC to the rect- 
angle AC, CB; 

therefore the square on AC is incommensurable with the rect- 
angle AC, CB, [x. 11] 

But the squares on AC, CB are commensurable with the 
square on AC ; for AC, CB are commensurable in square. 

[X 15] 

And twice the rectangle AC, CB is commensurable with 
the rectangle AC, CB, [x. 6] 

Therefore the squares on AC, CB are also incommen- 
surable with twice the rectangle AC, CB. [x. 13] 

But EG is equal to the squares on A C, CB, 
and HK is equal to twice the rectangle AC, CB ; 
therefore EG is incommensurable with HK, 
so that EH is also incommensurable in length with HN. 

[vl. I, X. Il] 

And they are rational ; 
therefore EH, HN are rational straight lines commensurable 

in square only. 

But, if two rational straight lines commensurable in square 
only be added together, the whole is the irrational which is 
called binomial. [x. 3 6 ] 

Therefore EN is a binomial straight line divided at H. 

In the same way EM, MN will also be proved to be 
rational straight lines commensurable in square only ; 
and EN will be a binomial straight line divided at different 
points, H and M. 



x. 44] PROPOSITION 44 97 

And EH is not the same with MN. 

For the squares on AC, CB are greater than the squares 
on AD, DB. 

But the squares on AD, DB are greater than twice the 
rectangle AD, DB ; 

therefore also the squares on AC, CB, that is, EG, are much 
greater than twice the rectangle AD, DB, that is, MK, 
so that EH is also greater than MN. 

Therefore EH is not the same with MN. 

Q. £. D. 

As the irrationality of the second bimedial straight line [x. 38] is proved by 
means of the irrationality of the binomial straight line [x. 36], so the present 
theorem is reduced to that of x. 43. 

Suppose, if possible, that the second bimedial straight line can be divided 
into its terms as such in two ways, i.e. that 

x +_y = li +y, 
where *', / are nearer equality than x, y. 

Apply x 1 +_v", 2xy to a rational straight line a-, i.e. let 
x* +_v* = <r», 
xxy — <rv. 

Then, as in x. 38, the areas x?+y, ixy are medial, so that au, ov are 
medial; 
therefore u, V are both rational and u a (t). 

Again, by hypothesis, x, y are medial straight lines commensurable in 
square only ; 
thetefore x^y. 

Hence x*vxy. 

And x 1 " (** +_v*), while xytxy; 
therefore («* -+y*) ~ txy, 
or <r» v <tv, 
and hence u*> v (1). 

Therefore, by (1) and (1), u, v are rational straight lines commensurable 
in square only; 

therefore u + v is a binomial straight line- 
Similarly, if x" ty* = w' and ix'y = <rv', 
u + v' will be proved to be a binomial straight line. 

And, since {x +_y)* = (*' +>')', and therefore (« + ») = («' + v'), it follows that 
a binomial straight line is divided as such in two ways : 
which is impossible. [x. 42] 

Therefore x + y, the given second bimedial straight line, can only be so 
divided in one way. 

In order to prove that u + v, u + v' represent a different division of the 
same straight line, Euclid assumes that x'+y t > ixy. This is of course an 
easy inference from 11. 7; but the assumption of it here renders it probable 
that the Lemma after x. 59 is interpolated. 



98 BOOK X fx. 45 



Proposition 45. 

A major straight line is divided at one and the same point 
only. 

Let AB be a major straight line divided at C, so that 
AC, CB are incommensurable in 

square and make the sum of the j 1 < B 

squares on AC, CB rational, but the 

rectangle AC, CB medial ; [x. 39] 

I say that AB is not so divided at another point. 

For, if possible, let it be divided at D also, so that AD, 
DB are also incommensurable in square and make the sum 
of the squares on AD, DB rational, but the rectangle con- 
tained by them medial. 

Then, since that by which the squares on AC, CB differ 
from the squares on AD, DB is also that by which twice the 
rectangle AD, DB differs from twice the rectangle AC, CB, 
while the squares on AC, CB exceed the squares on AD, 
DB by a rational area — for both are rational — 
therefore twice the rectangle AD, DB also exceeds twice the 
rectangle AC, CB by a rational area, though they are medial : 
which is impossible. [x. *6] 

Therefore a major straight line is not divided at different 
points; 

therefore it is only divided at one and the same point. 

Q. E. D. 

If possible, let the major irrational straight line be divided into terms in 
two ways, viz. as (i +y) and (x +y), where x 1 , y are supposed to be nearer 
equality than x, y. 

We have then, as in x. 42, 43, 

{x* +y) - (?P +/») = 2 *y - xxy. 

But, by hypothesis, {x*+y), (x'+y") are both rational, so that their 
difference is rational. 

Also, by hypothesis, 2x'y, xxy are both medial areas ; 
therefore the difference of two medial areas is a rational area : 
which is impossible. [x. 26] 

Therefore etc. 



x. 46, 47] PROPOSITIONS 45— 47 99 

Proposition 46. 

The side of a rational plus a medial area is divided at one 
point only. 

Let AB be the side of a rational plus a medial area 
divided at C, so that AC, CB are 

incommensurable in square and make a d~c B 

the sum of the squares on AC, CB 

medial, but twice the rectangle A C, CB rational ; [x. 40] 

I say that AB is not so divided at another point. 

For, if possible, let it be divided at D also, so that AD, 
DB are also incommensurable in square and make the sum 
of the squares on AD, DB medial, but twice the rectangle 
AD, DB rational. 

Since then that by which twice the rectangle AC, CB 
differs from twice the rectangle AD, DB is also that by 
which the squares on AD, DB differ from the squares on 

AC, CB, 

while twice the rectangle AC, CB exceeds twice the rectangle 

AD, DB by a rational area, 

therefore the squares on AD, DB also exceed the squares 
on AC, CB by a rational area, though they are medial : 
which is impossible. [x. 26] 

Therefore the side of a rational plus a medial area is not 
divided at different points ; 
therefore it is divided at one point only. 

Q. E. D. 

Here, as before, if we use the same notation, 

{*■ 4 f) - (x^ +/*) = 2X'/ - 2xy, 
and the areas on the left side are, by hypothesis, both media), while the areas 
on the right side are both rational. 

Thus the result of x. 26 is contradicted, as before. 

Therefore etc. 

Proposition 47. 

The side of the sum of two medial areas is divided at one 

point only. 

Let AB be divided at C, so that AC, CB are incommen- 
surable in square and make the sum of the squares on AC, 



IOO 



BOOK X 



[*. 47 



CB medial, and the rectangle AC, CB medial and also in- 
commensurable with the sum of the squares on them ; 
I say that AB is not divided at another point so as to fulfil 
the given conditions, 

M H 




to 






For, if possible, let it be divided at D, so that again AC 
is of course not the same asBD, but AC is supposed greater; 
let a rational straight line EF be set out, 
and let there be applied to EF the rectangle EG equal to the 
squares on AC, CB, 

and the rectangle HK equal to twice the rectangle AC, CB ; 
therefore the whole EK is equal to the square on AB. |n. 4] 

Again, let EL, equal to the squares on AD, DB, be applied 
to EF; 

therefore the remainder, twice the rectangle .^fZ), DB, is equal 
to the remainder MK. 

And since, by hypothesis, the sum of the squares on AC, 
CB is medial, 
therefore EG is also medial. 

And it is applied to the rational straight line EF; 
therefore HE is rational and incommensurable in length with 
EF. [x. 22] 

For the same reason 
HN is also rational and incommensurable in length with EF. 

And, since the sum of the squares on AC, CB is incom- 
mensurable with twice the rectangle AC, CB, 
therefore EG is also incommensurable with GN, 
so that EH is also incommensurable with HN. [vi. i, x. n] 

And they are rational ; 



x. 47] PROPOSITION 47, DEFINITIONS II 101 

therefore EH, HN are rational straight lines commensurable 

in square only ; 

therefore EM is a binomial straight line divided at H. [x. 36] 

Similarly we can prove that it is also divided at M. 

And EH is not the same with MM ; 
therefore a binomial has been divided at different points : 
which is absurd. [x. 43] 

Therefore a side of the sum of two medial areas is not 
divided at different points ; 
therefore it is divided at one point only. 

Using the same notation as in the note on x. 44, we suppose that, if 
possible, 

x+y = x' +/, 
and we put 

2xy = <rv ) zxy -trv J 

Then, since *' +y\ ixy are medial areas, and <r rational, 

u, v are both rational and w a (1). 

Also, by hypothesis, x*+y*^ xxy, 
whence u-jV ( 2 ), 

Therefore, by (i) and (*), u, v are rational and <•*-, 

Hence u + v is a binomial straight line. fx. 36] 

Similarly u + ©' is a binomial straight line. 

But u + v=u'+v'; 

therefore a binomial straight line is divided into terms in two ways : 
which is impossible. [x. 41] 

Therefore etc. 



DEFINITIONS II. 

1. Given a rational straight line and a binomial, divided 
into its terms, such that the square on the greater term is 
greater than the square on the lesser by the square on a 
straight line commensurable in length with the greater, then, 
if the greater term be commensurable in length with the 
rational straight line set out, let the whole be called a first 
binomial straight line; 

2. but if the lesser term be commensurable in length 
with the rational straight line set out, let the whole be called 
a second binomial ; 



loa BOOK X [Deff. ii. 3-6, x. 48 

3. and if neither of the terms be commensurable in length 
with the rational straight line set out, let the whole be called 
a third binomial. 

4. Again, if the square on the greater term be greater 
than the square on the lesser by the square on a straight line 
incommensurable in length with the greater, then, if the 
greater term be commensurable in length with the rational 
straight line set out, let the whole be called a fourth 
binomial ; 

5. if the lesser, a fifth binomial ; 

6. and if neither, a sixth binomial. 



Proposition 48. 
To find the first binomial straight line. 

Let two numbers AC, CB be set out such that the sum 
of them AB has to BC the ratio 
which a square number has to a D— H 

square number, but has not to CA * 

the ratio which a square number 

has to a square number ; a c b 

[Lemma 1 after x. 38] 
let any rational straight line D be set out, and let EF be 
commensurable in length with D t 

Therefore EF is also rational. 

Let it be contrived that, 
as the number BA is to AC, so is the square on EF to the 
square on FG. [x. 6, Pot.] 

But AB has to AC the ratio which a number has to a 
number ; 

therefore the square on EF also has to the square on FG 
the ratio which a number has to a number, 
so that the square on EF is commensurable with the square 
on FG. [x, 6] 

And EF is rational ; 
therefore FG is also rational. 

And, since BA has not to AC the ratio which a square 
number has to a square number. 



x. 4 8] DEFINITIONS II., PROPOSITION 48 103 

neither, therefore, has the square on EF to the square on FG 
the ratio which a square number has to a square number ; 
therefore EF is incommensurable in length with FG. [x. 9] 
Therefore EF, FG are rational straight lines commen- 
surable in square only ; 
therefore EG is binomial [x. 36] 

I say that it is also a first binomial straight line. 

For since, as the number BA is to A C, so is the square 
on EF to the square on FG, 
while BA is greater than AC, 

therefore the square on EF is also greater than the square 
on FG. 

Let then the squares on FG, H be equal to the square on 
EF. 

Now since, as BA is to A C, so is the square on EF to the 
square on FG, 
therefore, convertendo, 

as AB is to BC, so is the square on EF to the square on H. 

[v. 19, Por.] 

But AB has to BC the ratio which a square number has 
to a square number ; 

therefore the square on EF also has to the square on H the 
ratio which a square number has to a square number. 

Therefore EF is commensurable in length with H '; [x. 9] 
therefore the square on EF is greater than the square on FG 
by the square on a straight Hne commensurable with EF. 

And EF, FG are rational, and EF is commensurable in 
length with D. 

Therefore EF is a first binomial straight line. 

Q. E. D. 

Let kp be a straight line commensurable in length with p, a given rational 
straight line. 

The two numbers taken may be written p («* - it 1 ), /J* 1 , where {m'-n 1 ) is 
not a square. 

Take x such that 

pm i :p(m i - n<) = #? :x> (1), 

whence x = kp . 

* 

Then ip + x, or kp + kp ■ , is & first binomial straight li ne {2). 



104 



BOOK X 



[x. 48, 49 



To prove this we have, from (i), 

and x is rational, but Xvkp; 

that is, x is rational and "- kp, 

SO that kp -t x is a binomial straight line. 

Also, £*/)' being greater than a?, suppose Pp 1 - x* =y. 

Then, from ( 1 ), pm* : pt? = **/>' : >", 

whence^ is rational and * kp. 

Therefore kp + x is a jfw/ binomial straight line [x. DefT. 11. 1]. 

This binomial straight line may be written thus, 
kp + kp *J 1 - X'. 

When we come to x, 85, we shall find that the corresponding straight line 
with a negative sign is Hog first apotome, 

kp-kp>j\^W. 
Consider now the equation of which these two expressions are the roots. 
The equation is 

x 1 - ikp . x + A 5 £V = o- 
In other words, the first binomial and the first apotome correspond to the 
roots of the equation 

JC* — 20JC + X'o s = o, 
where a = kp. 






Proposition 49. 
To find the second binomial straight line. 

Let two numbers AC, CB be set out such that the sum 
of them AB has to BC the ratio which 
a square number has to a square number, 
but has not to AC the ratio which a 
square number has to a square number ; 
let a rational straight line D be set out, 
and let EF be commensurable in length 
with D ; 
therefore EF is rational. 

Let it be contrived then that, 
as the number CA is to AB, so also is the square on EF to 
the square on FG ; [x. 6, Por.] 

therefore the square on EF is commensurable with the square 
on FG. [x. 6] 

Therefore FG is also rational. 

Now, since the number CA has not to AB the ratio which 
a square number has to a square number, neither has the 



x. 49] PROPOSITIONS 48, 49 i°5 

square on EF to the square on FG the ratio which a square 
number has to a square number. 

Therefore EF is incommensurable in length with FG \ 

[*-9] 
therefore EF, FG are rational straight lines commensurable 
in square only ; 
therefore EG is binomial. [x. 36] 

It is next to be proved that it is also a second binomial 
straight line. 

For since, inversely, as the number BA is to AC, so is 
the square on GF to the square on FE, 
while BA is greater than AC, 
therefore the square on GF is greater than the square on FE. 

Let the squares on EF, H be equal to the square on GF; 

therefore, convertendo, as AB is to BC, so is the square on 
FG to the square on H. [v. 19, Por.] 

But AB has to BC the ratio which a square number has 
to a square number ; 

therefore the square on FG also has to the square on H the 
ratio which a square number has to a square number. 

Therefore FG is commensurable in length with H ; [x. 9] 
so that the square on FG is greater than the square on FE 
by the square on a straight line commensurable with FG. 

And FG, FE are rational straight lines commensurable 
in square only, and EF, the lesser term, is commensurable in 
length with the rational straight line D set out. 

Therefore EG is a second binomial straight line. 

Q. E. D. 

Taking a rational straight line kp commensurable in length with p, and 
selecting numbers of the same form as before, viz. p (m* - re 1 ), /« J , we put 
fi{m i -tt') :/>m , = ¥p i : & (1), 



so that x = kp 



Vm*-«' 



= kp .-1— , say (a). 

•/I —A* 



Just as before, x is rational and "- kp, 
whence kp + * is a binomial straight line. 
By(i) f x 1 >k , p\ 



io« BOOK X [x, 49, 50 

Let **-iy=y ( 

whence, from (r), fm 1 :pn , = x 1 :y*, 

and y is therefore rational and « x. 

The greater term of the binomial straight line is x and the lesser kp, and 

Ji -A' 
satisfies the definition of the second binomial straight tine. 
The corresponding second apotome [x. 86] is 

-£= -k P . 

The equation of which the two expressions are the roots is 

or 
where 



Proposition 50. 
To find the third binomial straight line. 

Let two numbers AC, CB be set out such that the sum 
of them AB has to BC the ratio which a square number has 
to a square number, but has not to AC the ratio which a square 
number has to a square number. 




=^ ° 



Let any other number D, not square, be set out also, and 
let it not have to either of the numbers BA. AC the ratio 
which a square number has to a square number. 

Let any rational straight line E be set out, 

and let it be contrived that, as D is to AB, so is the square 
on E to the square on FG ; [x. 6, Por,] 

therefore the square on E is commensurable with the square 
on FG. [x. 6] 

And E is rational ; 

therefore FG is also rational. 



x. $o] PROPOSITIONS 49, 50 107 

And, since D has not to AB the ratio which a square 
number has to a square number, 

neither has the square on E to the square on FG the ratio 
which a square number has to a square number ; 
therefore E is incommensurable in length with FG. [x. 9] 

Next let it be contrived that, as the number BA is to AC, 
so is the square on FG to the square on GH; [x. 6, Por.] 

therefore the square on FG is commensurable with the square 
on GH, [x. 6] 

But FG is rational ; 
therefore GH is also rational. 

And, since BA has not to A C the ratio which a square 
number has to a square number, 

neither has the square on FG to the square on HG the ratio 
which a square number has to a square number ; 
therefore FG is incommensurable in length with GH. [x. 9] 

Therefore FG, GH are rational straight lines commen- 
surable in square only ; 
therefore FH is binomial. [x. 36] 

I say next that it is also a third binomial straight line. 

For since, as D is to AB, so is the square on E to the 
square on FG, 

and, as BA is to AC, so is the square on FG to the square 
on GH, 

therefore, ex aequali, as D is to A C, so is the square on E to 
the square on GH. [v. z*] 

But D has not to AC the ratio which a square number 
has to a square number; 

therefore neither has the square on E to the square on GH 
the ratio which a square number has to a square number ; 
therefore E is incommensurable in length with GH. [x. 9] 

And since, as BA is to AC, so is the square on FG to 
the square on GH, 
therefore the square on FG is greater than the square on GH. 

Let then the squares on GH, K be equal to the square 
on FG ; 



10* BOOK X [x. s 



o 



therefore, converlendo, asAB is to BC, so is the square on FG 
to the square on K. [v. 19, Por.] 

But AB has to BC the ratio which a square number has 
to a square number ; 

therefore the square on FG also has to the square on K the 
ratio which a square number has to a square number ; 
therefore FG is commensurable in length with K. [x. 9] 

Therefore the square on FG is greater than the square on 
GH by the square on a straight line commensurable with FG. 

And FG, GH are rational straight lines commensurable 
in square only, and neither of them is commensurable in length 
with E. 

Therefore FH is a third binomial straight line. 

Q. E. D. 

Let p be a rational straight line. 

Take the numbers q (»' — «"), qn 1 , 
and let p be a third number which is not a square and which has not to qm 1 
or q (m* - «') the ratio of square to square. 

Take x such that / : qm' = p' : x* (i), 

Thus x is rational and « p (2). 

Next suppose that qm' :?(*»* — «*) ~x? -,y* (3). 

It follows that y is rational and «- * (4), 

Thus (x +y) is a binomial straight line. 

Again, from (1) and {3), ex aequaU, 

/;q(m'-«') = p>;y> (5), 

whence y^p (6). 

Suppose that x? -f = **. 

Then, from {3), eonverUndo, 

qm 1 : qn % m x? : i 1 , 
whence £ rt * 

Thus ■Jx*-y>"x, 

and x, y are both w p ; 
therefore 1 +_c is a third binomial straight line. 

Now, from (i), x = p. — JO-i 

.... - Jm* -n*. Jq 

and, by (5), y = p. — "-£ - — ■ 

Thus the third binomial is 



A 



^p(m+ Jm* - «")» 

which we may write in the form 

m */& . p + m Jh . p */i - X*. 



x. jo, 51] PROPOSITIONS so, Si to 9 

The corresponding third apotomt [x. 87] is 

m Jk . p — m JJt . p Vi - A*. 
The two expressions are accordingly the roots of the equation 
** - in Jk . px + AWiip" = o, 
or x* — tax + X'o* - o, 

where a = m „JA . p. 

See also note on x. 53 {ad fin.). 



Proposition 51.' 

To find I he fourth binomial straight line. 

Let two numbers AC, CB be set out such that AB 
neither has to BC, nor yet to AC, the ratio 
which a square number has to a square number. 

Let a rational straight line O be set out, 
and let EF be commensurable in length with D ; 
therefore EF is also rational. 

Let it be contrived that, as the number BA 
is to AC, so is the square on EF to the square 
on FG ; [x, 6, For,] 

therefore the square on EF is commensurable 
with the square on FG ; [x. 6] 

therefore FG is also rational. 

Now, since BA has not to AC the ratio which a square 
number has to a square number, 

neither has the square on EF to the square on FG the ratio 
which a square number has to a square number ; 
therefore EF is incommensurable in length with FG. [x. 9] 

Therefore EF, FG are rational straight lines commen- 
surable in square only ; 
so that EG is binomial. 

I say next that it is also a fourth binomial straight line. 

For since, as BA is to AC, so is the square on EF to the 
square on FG, 
therefore the square on EF is greater than the square on FG. 

Let then the squares on FG, H be equal to the square 
on EF; 



no BOOK X [x>S*»$> 

therefore, convertendo, as the number AB is to BC, so is the 
square on EF to the square on H. [v. 19, Por.] 

But AB has not to BC the ratio which a square number 
has to a square number ; 

therefore neither has the square on EF to the square on H 
the ratio which a square number has to a square number. 

Therefore EF is incommensurable in length with H \ [x. 9] 
therefore the square on EF is greater than the square on GF 
by the square on a straight line incommensurable with EF. 

And EF, FG are rational straight lines commensurable in 
square only, and EF is commensurable in length with D. 

Therefore EG is a fourth binomial straight line. 

Q. E. D. 

Take numbers m, n such that (m + n) has not to either m or n the ratio of 

square to square. 

Take x such that <« + »):« = Ay ; x*, 

I 

whence i=A)»/ 

kp 






[ + A 



say. 



kp 
Then kp + x, or kp + ■ , is s. fourth binomial straight line. 

For J&p'-x? is incommensurable in length with kp, and kp is com- 
mensurable in length with p. 

The corresponding/twrr'ii afotome [x, 88] Is 

•Ji + K 
The equation of which the two expressions are the loots is 

X*-2kp.X+j-^k>p'=0, 

Or X*— 2MT + - rB 1 =0, 

I + A ^ 
where a=kp. 

Proposition 52. 

To find the fifth binomial straight line. 

Let two numbers AC, CB be set out such that AB has 
not to either of them the ratio which a square number has 
to a square number ; 
let any rational straight line D be set out, 



x. si] PROPOSITIONS si, $3 in 

and let EF be commensurable with D ; 
therefore EF is rational. 

Let it be contrived that, as CA is to AB, so is the 
square on EF to the square on FG. [x. 6, Par.] 

But CA has not to AB the ratio which a 
square number has to a square number ; 
therefore neither has the square on EF to the 
square on FG the ratio which a square number 
has to a square number. 

Therefore EF, FG are rational straight 
lines commensurable in square only ; [* 9] 
therefore EG is binomial. [x. 36] 



X 



H Q 



t say next that it is also a fifth binomial straight line. 

For since, as CA is to AB, so is the square on EF tc 
the square on FG, 

inversely, as BA is to AC, so is the square on FG to the 
square on FE ; 

therefore the square on GF is greater than the square on FE. 

Let then the squares on EF, H be equal to the square 
on GF; 

therefore, conventendo, as the number AB is to BC, so is the 
square on GF to the square on H. [v. 19, Por.] 

But AB has not to BC the ratio which a square number 

has to a square number ; 

therefore neither has the square on EG to the square on H 
the ratio which a square number has to a square number. 

Therefore FG is incommensurable in length with H \ [x. 9] 

so that the square on FG is greater than the square on FE 
by the square on a straight line incommensurable with FG. 

And GF, FE are rational straight lines commensurable 
in square only, and the lesser term EF is commensurable in 
length with the rational straight line D set out. 

Therefore EG is a fifth binomial straight line. 

Q. E. D. 

If m, n be numbers of the kind taken in the last proposition, take x such 
that 

m : (m + ri) = ifp' : srt 



113 



In this case 



BOOK X 

= kp Jt + A, say, 



[*• 5*. S3 



and x =- kp. 

Then ip Vi + A + Ap is ay£#A binomial straight line. 

For <Jx* — &p\ or ^\.kp, is incommensurable in length with Up si + A, 
or #; 

and £p, but not £p Vi + A, is commensurable in length with p. 

The corresponding^/* apotome [x. 89] is 

>&/>^i +X-V 

The equation of which the fifth binomial and the fifth apotome are the 
roots is 



or 

where 



x* — ikp <J\ + A . x + Av5y = o, 

ar — 2«x + r o = o, 

1 + A 

a ■ kp V I + A* 



Proposition 53. 
To find the sixth binomial straight line. 

Let two numbers AC, CB be set out such that AB has 
not to either of them the ratio which a 
square number has to a square number ; 
and let there also be another number D 
which is not square and which has not to 
either of the numbers BA, AC the ratio 
which a square number has to a square 
number. 

Let any rational straight line E be set 
out, 

and let it be contrived that, as D is to AB, 
so is the square on E to the square on EG ; [x. 6, Por.) 

therefore the square on E is commensurable with the square 
on EG. [x. 6] 

And E is rational ; 

therefore EG is also rational. 

Now, since D has not to AB the ratio which a square 
number has to a square number, 



x. S3] PROPOSITIONS 5*, 53 113 

neither has the square on E to the square on FG the ratio 

which a square number has to a square number ; 

therefore E is incommensurable in length with FG, [x. 9] 

Again, let it be contrived that, as BA is to AC, so is the 
square on FG to the square on GH. [x. 6, Por.] 

Therefore the square on FG is commensurable with the 
square on HG, [x. 6] 

Therefore the square on HG is rational ; 
therefore HG is rational. 

And, since BA has not to AC the ratio which a square 
number has to a square number, 

neither has the square on FG to the square on GH the ratio 
which a square number has to a square number ; 
therefore FG is incommensurable in length with GH. [x. 9] 

Therefore FG, GH are rational straight lines commen- 
surable in square only ; 
therefore FH is binomial. [x. 36] 

It is next to be proved that it is also a sixth binomial 
straight line. 

For since, as D is to AB, so is the square on E to the 
square on FG, 

and also, as BA is to AC, so is the square on FG to the 
square on GH, 

therefore, ex aequaii, as D is to AC, so is the square on E 
to the square on GH. [v. 12] 

But D has not to AC the ratio which a square number 
has to a square number ; 

therefore neither has the square on E to the square on GH 
the ratio which a square number has to a square number ; 
therefore E is incommensurable in length with GH. [x. 9] 

But it was also proved incommensurable with FG ; 
therefore each of the straight lines FG, GH is incommen- 
surable in length with E. 

And, since, as BA is to AC, so is the square on FG to 
the square on GH, 
therefore the square on FG is greater than the square on GH. 

Let then the squares on GH, K be equal to the square 
on FG; 



H4 BOOK X [x. 53 

therefore, convertendo, as AB is to BC, so is the square on FG 
to the square on K. [v. iy, For.] 

But AB has not to BC the ratio which a square number 
has to a square number ; 

so that neither has the square on FG to the square on K the 
ratio which a square number has to a square number. 

Therefore FG is incommensurable in length with K; [x. 9] 
therefore the square on FG is greater than the square on GH 
by the square on a straight line incommensurable with FG. 

hx\AFG t GHaxz. rational straight lines commensurable in 
square only, and neither of them is commensurable in length 
with the rational straight line E set out. 

Therefore FN is a sixth binomial straight line. 

Q. E. D. 

Take numbers m, n such that (m + 11) has not to either of the numbers 
m, n the ratio of square to square ; take also a third number /, which is not 
square, and which has not to either of the numbers (m + «), m the ratio of 
square to square. 

Let /> : (»t + n) = p": a* (1) 

and {m + ») : m- x* : y* •••(*)■ 

Then shall (x +y) be a sixth binomial straight line. 

For, by (1), x is rational and <j p. 
By (2), since x is rational, 

y is rational and u x. 

Hence x, y are rational and commensurable in square only, so that (x +y) 
is a binomial straight line. 

Again, ex aequali, from (1) and (z), 

t :m = p*:y % (3). 

whence y u p. 

Thus x, y are both incommensurable in length with p. 
Lastly, from (a), comtrftndo, 

(m + ») : n = x* ; (x' — y 1 ), 
so that \fx* — y' u jr. 

Therefore (x + y) is a sixth binomial straight line. 
Now, from {1) and (3), 

■n 

t(tM say, 



ym + 
"7 



/■J*- */j=P>J*i say. 

and the «M5i trinomial straight line may be written 

Jh . p +■ jk . p. 

The corresponding sixth apotomt is [x. 90] 

«jk.p-J\.p- } 



x. 53. Lemma] PROPOSITION 53 n S 

and the equation of which the two expressions are the toots is 

x* - a Jk . px + (i — A) p* » o, 

*-* , 
or ar — 2<l£ + — j— o = o, 

ft 

where o = Jk . p. 

Tannery remarks ("De la solution geometrique des problemes du second 
degr£ avant Euclide" in Mimoires de fa Sociitides sciences physiques el naturdies 
it Bordeaux, 2= Serie, T. iv.) that Euclid admits as binomials and apotomes 
the third and sixth binomials and apotomes which are the square roots of first 
binomials and apotomes respectively. Hence the third and sixth binomials 
and apotomes are the positive roots of biquadratic equations of the same form 
as the quadratics which give as roots the first and fourth binomials and 
apotomes. But this remark seems to be of no value because (as was pointed 
out a hundred years ago by Cossali, 11. p. 160) the squares of ail the six 
binomials and apotomes (including the first and fourth) give first binomials 
and apotomes respectively. Hence we may equally welt regard them all as 
roots of biquadratics reducible to quadratics, or generally as roots of equations 
of the form 

•**■+ aa. a*"" 1 ±^ = 0; 
and nothing is gained by raising the degree of the equations in this way. 

It is, of course, easy to see that the most general form of binomial and 
apotome, viz. 

p . ^h ± p . ,/A, 
give first binomials and apotomes when squared. 

For the square is p{(A + A)p + s */hk . p\ ; and the expression within the 
bracket is a first binomial or apotome, because 

(1) k~k>?jkK, 

(2) V{A + A)*-+iA = k - A, which is « (A + A), 

(3) (* + *)p"f- 

Lemma. 

Let there be two squares AB, BC, and let them be placed 
so that DB is in a straight line with BE ; 
therefore FB is also in a straight line with 
BG. 

Let the parallelogram AC be completed; 
I say that AC is a square, that DG is a 
mean proportional between AB, BC, and 
further that DC is a mean proportional 
between AC, CB. 

For, since DB is equal to BF, and BE to BG, 
therefore the whole DE is equal to the whole FG. 

But DE is equal to each of the straight lines AH, KC, 
and FG is equal to each of the straight lines AK, HC ; [1. 34] 






B 









u6 BOOK X [Lemma, x. 54 

therefore each of the straight lines AH, KC is also equal to 
each of the straight lines AK, HC. 

Therefore the parallelogram A C is equilateral. 

And it is also rectangular ; 
therefore AC is a square. 

And since, as FB is to BG, so is DB to BE, 
while, as FB is to BG, so is AB to DG, 
and, as DB is to BE, so is DG to BC, [vi. i] 

therefore also, as AB is to DG, so is DG to i?C. [v. 1 1] 

Therefore DG is a mean proportional between -<4Z?, Z?C 

I say next that DC is also a mean proportional between 
AC, CB. 

For since, as AD is to DK, so is KG to Ct — 
for they are equal respectively — 

and, componendo, as AK is to KD, so is KC to CC, [v. 18] 
while, as AK is to KD, so is ^4 C to CM 
and, as KC is to CG, so is DC to CZ?, [vi. i] 

therefore also, as AC is to DC, so is DC to i?C [v. n] 

Therefore DC is a mean proportional between AC, CB. 
Being what it was proposed to prove. 

It is here proved that 

x* :xy-xy -.y, 

and (* +>)* : (x +y)y = (* + y)y -.y. 

The first of the two results is proved in the course of x. 25 (lines 6—8 on 
p. 57 above). This fact may, I think, suggest doubt as to the genuineness 
of this Lemma. 

Proposition 54. 

If an area be contained by a rational straight line and the 
first binomial, the "side" of the area is the irrational straight 
line which is called binomial. 

For let the area A C be contained by the rational straight 
line AB and the first binomial AD ; 

I say that the "side" of the area AC is the irrational straight 
line which is called binomial. 

For, since AD is a first binomial straight line, let it be 
divided into its terms at E, 
and let AE be the greater term. 



X.§4] 



PROPOSITION 54 



iij 



It is then manifest that AE, ED are rational straight lines 
commensurable in square only, 

the square on AE is greater than the square on ED by the 
square on a straight line commensurable with AE, 
and AE is commensurable in length with the rational straight 
line A3 set out [x. Deff. h. i] 

Let ED be bisected at the point F. 

A G E f d r a 



H K 



M 







N 


P 



Then, since the square on AE is greater than the square 
on ED by the square on a straight line commensurable with 
AE, 

therefore, if there be applied to the greater AE a. parallelogram 
equal to the fourth part of the square on the less, that is, to 
the square on EF, and deficient by a square figure, it divides 
it into commensurable parts. [x. 17] 

Let then the rectangle AG, GE equal to the square on 
EF be applied to AE ; 
therefore AG is commensurable in length with EG. 

Let GH, EK, FL be drawn from G, E, F parallel to 
either of the straight lines AE, CD ; 

let the square SN be constructed equal to the parallelogram 
AH, and the square NQ equal to GK, [11. 14] 

and let them be placed so that MN is in a straight line with 
NO; 
therefore RN is also in a straight line with NP. 

And let the parallelogram SQ be completed ; 
therefore SQ is a square. [Lemma] 

Now, since the rectangle AG, GE is equal to the square 
on EF, 

therefore, as A G is to EF, so is FE to EG ; [vi. 1 7) 

therefore also, as AH is to EL, so is EL to KG ; [vi. 1] 

therefore EL is a mean proportional between AH, GK. 

But A His equal to SN, and GK to NQ ; 
therefore EL is a mean proportional between SN, NQ. 



nS BOOK X {x. 54 

But MR is also a mean proportional between the same 
SN, NQ; [Lemma] 

therefore EL is equal to MR, 
so that it is also equal to PO. 

But AH, GK are also equal to SN, NQ ; 
therefore the whole AC is equal to the whole SQ, that is, to 
the square on MO ; 
therefore MO is the "side" of AC. 

I say next that MO is binomial. 

For, since AG is commensurable with GE, 
therefore AE is also commensurable with each of the straight 
lines AG, GE. [x. 15] 

But A E is also, by hypothesis, commensurable with AB ; 
therefore AG, GE are also commensurable with AB. [x. it] 

And AB is rational ; 
therefore each of the straight lines AG, GE is also rational ; 
therefore each of the rectangles AH, GK is rational, [x. 19] 
and AH is commensurable with GK. 

But AH is equal to SN, and GK to NQ ; 
therefore SN, NQ, that is, the squares on MN, NO, are 
rational and commensurable. 

And, since AE is incommensurable in length with ED, 
while AE is commensurable with AG, and DE is commen- 
surable with EF, 

therefore AG is also incommensurable with EF, 
so that AH is also incommensurable with EL. [vi. 

But W// is equal to S/V, and EL to AW ; 
therefore SN is also incommensurable with MR. 

But, as SAf is to MR, so is PN to A'j? ; 
therefore PN is incommensurable with ATtf. 

But PN is equal to MN, and A>7? to A^ ; 
therefore MN is incommensurable with NO. 

And the square on MN is commensurable with the square 
on NO, 

and each is rational ; 

therefore MN, NO are rational straight lines commensurable 
in square only. 



[X. 


'3] 


I,X, 


»] 


[vi 


,. 1] 


[X. 


»] 



x. 54] PROPOSITION 54 119 

Therefore MO is binomial [x. 36] and the "side" of AC, 

Q. E. D. 

1. "side," I us? the word " side " in the sense explained in the note on x. Def. 4 
(p. 13 above), i.e. as short for "side or a square equal to." The Greek is ^ ri xt^er 

A first binomial straight line being, as we have seen in x. 48, of the form 
ip + kpji^k 1 , 
the problem solved in this proposition is the equivalent of finding the square 
root of this expression multiplied by p, or of 

p(hp + kp-Ji -A 1 ), 

and of proving that the said square root represents a binomial straight line 
as defined in x. 36. 

The geometrical method corresponds sufficiently closely to the algebraical 
one which we should use. 
First solve the equations 

u + v^kp I „ 

iw=1*V(i-A') I K * h 

Then, if u, v represent the straight lines so found, put 

x* = pu 

y 

and the straight line (x +y) is the square root required. 
The actual algebraical solution of (1) gives 

u - V = kp . X, 
so that u - \kp (1 + X), 

v = \kp(t -X), 

n 

and therefore * = P*/-( I + A), 



':%} »■ 



>~*s/\ 



'(I-*). 



and 



*+^ = P^/-( I + X ) + Py -(i-*)- 



This is clearly a binomial straight line as defined in x. 36. 
Since Euclid has to express his results by straight lines in his figure, and 
has no symbols to make the result obvious by inspecti on, he is obliged to 

Strove (1) that (* + /) is the square root of p(kp + kp >J 1 - A*), and (1) that 
x + y) is a binomial straight line, in the following manner. 
First, he proves, by means of the preceding Lemma, that 

qr«|/«/t=? (3); 

therefore (* *yf = x*+y + xxy 

= p{u + v) + ixy 
= V + kp* Vi - A», by (1) and {3), 

so that x+y^Jp(kp + kp>li- X'). 



no BOOK X [x. 54, 55 

Secondly, it results from (i), [by x. 17], that 

M n V, 

so that u, v are both « (u + v), and therefore « p (4); 

thus «, n are rational, 

whence pu, pv are both rational, and 

pu n pv. 

Therefore a?, ^v 1 are rational and commensurable (5). 

Next, kp yj kp 1/ 1 — if, 
and kp « u, while kp Vr - X" « £,*p -Ji-X'; 
therefore » ~ \kp J\ - X', 

whence pv ~ $kp' «/r -X*, 

or j^ u ay, 

so that x u_f. 

By this and (5), x, _y are rational and «-, so that (jr+.p) is a binomial 
straight line. [x. 36] 

X. 91 will prove in like manner that a like theorem holds for apotomes, 
viz. that 

P J\ (t * A)-* v/fo^*) = Vp(*P-Wi"-A*). 

Since the ^rri binomial straight line and the ./frrf apotomt are the root* of 
the equation 

jc 5 — zip . x + Wp* = o, 
this proposition and x. 91 give us the solution of the biquadratic equation 

X 4 -2kp'.X* + XWp' - o. 

Proposition 55. 

If an area be contained by a rational straight line and the 
second binomial, the "side" of the area is the irrational straight 
line vthich is called a first bimediat. 

For let the area ABCD be contained by the rational 
s straight line AB and the second binomial AD ; 

I say that the "side" of the area AC is a first bimedial straight 

line. 

For, since AD is a second binomial straight line, let it be 

divided into its terms at E, so that AE is the greater term ; 
10 therefore AE, ED are rational straight lines commensurable 

in square only, 

the square on AE is greater than the square on ED by the 

square on a straight line commensurable with AE, 

and the lesser term ED is commensurable in length with AB. 

[x. Deff. 11. 1] 
ij Let ED be bisected at E, 



x-S5] 



PROPOSITIONS 54, SS 



in 



and let there be applied to AE the rectangle AG, GE equal 
to the square on EF and deficient by a square figure ; 
therefore AG is commensurable in length with GE. [x. 17] 
Through G, £, F let GH, EK, FL be drawn parallel to 
*>AB, CD, 
let the square SN be constructed equal to the parallelogram 
AH, and the square NQ equal to GK, 
and let them be placed so that MN is in a straight line with 

N0 > 
is therefore RN is also in a straight line with NP. 

r Q 



g e 









N 







JO 



Let the square SQ be completed. 

It is then manifest from what was proved before that MR 
is a mean proportional between SN, NQ and is equal to EL, 
and that MO is the "side" of the area AC. 

It is now to be proved that MO is a first bimedial straight line. 

Since AE is incommensurable in length with ED, 
while ED is commensurable with AS, 
therefore AE is incommensurable with AB. [x. 13] 

And, since AG is commensurable with EG, 
35 AE is also commensurable with each of the straight lines 
AG, GE. (x. 15] 

But AE is incommensurable in length with AB ; 
therefore AG, GE are also incommensurable with AB. [x. 13] 

Therefore BA, AG and BA, GE are pairs of rational 
40 straight lines commensurable in square only ; 
so that each of the rectangles AH, GK is medial. [x. ai] 

Hence each of the squares SN, NQ is medial. 

Therefore MN, NO are also medial. 

And, since AG is commensurable in length with GE, 
45 AH is also commensurable with GK, [vi. 1, x. 11] 

that is, SN is commensurable with NQ, 
that is, the square on MN with the square on NO. 



l" BOOK X fx. 55 

And, since AE is incommensurable in length with ED, 

while AE is commensurable with AG, 
so and ED is commensurable with EF, 

therefore AG is incommensurable with EF; [x. 13] 

so that AH is also incommensurable with EL, 

that is, SN is incommensurable with MR, 

that is, PN with NR, [vi. i, x. n] 

ss that is, MN is incommensurable in length with NO. 

But MN, NO were proved to be both medial and com- 
mensurable in square ; 

therefore MN, NO are medial straight lines commensurable 
in square only. 

60 1 say next that they also contain a rational rectangle. 

For, since DE is, by hypothesis, commensurable with each 
of the straight lines AB, EF, 

therefore EF is also commensurable with EK. [x. u] 

And each of them is rational ; 

65 therefore EL, that is, MR is rational, L x - '9] 

and MR is the rectangle MN, NO. 

But, if two medial straight lines commensurable in square 

only and containing a rational rectangle be added together, the 

whole is irrational and is called a first bimedial straight line. 

[*■ 37] 
70 Therefore MO is a first bimedial straight line. 

Q. E. D. 

39. Therefore BA, AG and BA, OE are pairs of rational straight line a com- 
mensurable in square only. The text has "Therefore BA, AG, GE are rational straight 
lines commensurable in square only." which I have altered because it would naturally convey 
the impression that any two of the three straight lines ate com mensurable in square only, 
whereas AG, GE are commensurable in length {1. 18), and it is only the other two pain 
which are commensurable in square only. 

A second binomial straight tine being [x. 49] of the form 
the present proposition is equivalent to finding the square root of the expression 



x. 55] PROPOSITION 55 1*3 

As in the last proposition, Euclid finds u, o from the equations 
u + v = - 






then finds x, y from the equations 



x* = pu\ 



(1), 
-(a). 



and then proves (a) that 

*+J=vA (;/== + ^). 

and (/J) that (.v + _v) is a first bimedial straight line [x. 37]. 

The steps in the proof are as follows. 

For (a) reference to the corresponding part of the previous proposition 
suffices. 

09) By (!) and X. 17, 

«« w; 

therefore u, v are both rational and rt (« + f), and therefore */ f> [by (1 )]...{ 3). 

Hence pu, pv, or x*, y, are medial areas, 
so that x,y are also medial (4)- 

But, since u « p, 

^"^ (5)- 

Again (« + w), or 7==, yip, 

so that it u i^p , 

whence pa v lip', 

or x? v xy, 

and *^> (6). 

Thus [(4), (5), {6}] *, ,v are medial and "-. 
Lastly, acy = iip 1 , which is rational. 
Therefore (x + y) is & first bimedial straight line. 
The actual straight lines obtained from (1) are 

U=\-^=:kp ) 

vi-i 1 I 
1 >-* 1 I" 

sothat *+J' = " N /»(T^) + W^(tTx)' 

The corresponding jfrj/ apolome of a medial straight line found in x, 92 
being the same thing with a minus sign between the terms, the two expressions 
are the roots of the biquadratic 






being the equation in r 1 corresponding to that in x in x. 49. 



"4 



BOOK X 



fx.56 



Proposition 56. 

If an area be contained by a rational straight line and the 
third binomial, the "side" 0/ the area is the irrational straight 
line called a second bimedial. 

For let the area ABCD be contained by the rational 
straight line AB and the third binomial AD divided into its 
terms at E, of which terms AE is the greater ; 
I say that the **side" of the area AC is the irrational straight 
line called a second bimedial. 

For let the same construction be made as before. 

r a 



A Q E F C 


> 

M 

( 














ft 




B 1 


ri 1 


t 




1 P 









Now, since AD is a third binomial straight line, 
therefore AE, ED are rational straight lines commensurable 
in square only, 

the square on AE is greater than the square on ED by the 
square on a straight line commensurable with AE, 
and neither of the terms AE, ED is commensurable in length 
with AB. I*. Deff. 11. 3] 

Then, in manner similar to the foregoing, we shall prove 
that MO is the "side" of the area AC, 

and MN, NO are medial straight lines commensurable in 
square only ; 
so that MO is bimedial. 

It is next to be proved that it is also a second bimedial 
straight line. 

Since DE is incommensurable in length with AB, that is, 
with EK, 

and DE is commensurable with EE, 
therefore EE is incommensurable in length with EK. [x. 13] 

And they are rational ; 



x. s6, 57] PROPOSITIONS 56, 57 1*5 

therefore FE, EK are rational straight lines commensurable 
in square only. 

Therefore EL, that is, MR, is medial. [x. ji] 

And it is contained by MN, NO ; 
therefore the rectangle MN, NO is medial. 

Therefore MO is a second bi medial straight line. [x. 3 8 ] 

Q. E. D. 

This proposition in like manner is the equivalent of finding the square 
1 jot or the product of p and the third Hnomial [x. 50], i.e. of the expression 

pUk.p+JM.pJT^X). 
As before, put 

« + v=Jk.p 1 

uv = \kp l (i-k t ) J w ' 

Next, «, v being found, let 

x* = pu, 

then (a- + v) is the square root required and is a second bimedial straight line. 

t x -38] 
For, as in the last proposition, it is proved that (x +_v) is the square root, 
and x, y are medial and «-» 

Again, xy = § Jk . p" ,/i -A*, which is media/. 
Hence (* +_v) is a second bimedial straight line. 
By solving equations (1), we find 

*«1<V*. J. + *</*.*>), 

and x+y = p J ^ (. + X) + p J& {l - X). 

The corresponding second apotome of a medial found in X. 93 is the same 
thing with a minus sign between the terms, and the two are the roots (cf. note 
on x. 50) of the biquadratic equation 

x 4 -3ji.p*x> + *»V = »■ 



Proposition 57. 

If an area be contained by a rational straight line and the 
fourth binomial, the "side" of the area is the irrational straight 
line called major. 

For let the area AC he. contained by the rational straight 
line AB and the fourth binomial AD divided into its terms 
at E, of which terms let AE be the greater ; 
I say that the "side" of the area AC is the irrational straight 
line called major. 



n6 



BOOK X 






[x-57 



For, since AD is a fourth binomial straight line, 
therefore AE, ED are rational straight lines commensurable 
in square only, 

the square on AE is greater than the square on ED by the 
square on a straight line incommensurable with AE, 
and AE is commensurable in length with AB. [x. Deff. 11. 4] 

Let DE be bisected at F, 
and let there be applied to AE a parallelogram, the rectangle 
AG, GE, equal to the square on EF; 
therefore AG is incommensurable in length with GE, [x. 18] 

Let GH, EK, FL be drawn parallel to AB, 
and let the rest of the construction be as before ; 
it is then manifest that MO is the "side" of the area AC. 






A 


( 


3 E 


* 


D 














£ 


\ » 


\ > 


I 


C 







R 


a 


M 








N 






S 


1 F 


> 








It is next to be proved that MO is the irrational straight 
line called major. 

Since AG is incommensurable with EG, 
AH is also incommensurable with GK, that is. SN with NQ ; 

[VI. I, X. It] 

therefore MN, NO are incommensurable in square. 

And, since AE is commensurable with A3, 
AK is rational ; [x. 19] 

and it is equal to the squares on MN, NO ; 
therefore the sum of the squares on MN, NO is also rational. 

And, since DE is incommensurable in length with AB, 
that is, with EK, 

while DE is commensurable with EF, 
therefore EF is incommensurable in length with EK. [x. 13] 

Therefore EK, EF are rational straight lines commen- 
surable in square only; 
therefore LE, that is, MR, is medial. [x. 21] 

And it is contained by MN, NO ; 
therefore the rectangle MN, NO is medial. 



x. 57] PROPOSITION 57 1*7 

And the [sum] of the squares on MN, NO is rational, 
and MN, NO are incommensurable in square. 

But, if two straight lines incommensurable in square and 
making the sum of the squares on them rational, but the 
rectangle contained by them medial, be added together, the 
whole is irrational and is called major. [x. 39] 

Therefore MO is the irrational straight line called major 
and is the "side" of the area AC. Q. e. d. 

The problem here is to find the square root of the expression [cf. x. 51] 

\ -Ji + k/ 

The procedure is the same. 
Find », v from the equations 



u + v = kp \ 

W = l— pT f 

4 i + M 



-M. 

and, if x* = pu 1 , . 

y=pW *"* 

(x +y) is the required square root. 

To prove that {# +>) is the major irrational straight line Euclid argues 
thus. 

By x, 18, u w v, 

therefore pu w pv, 

or ^^y, 

sothat x^-y .... (3). 

Now, since (u + v)* p, 

(u + v)p, or (x'+y), is a rational area {4). 

An* 

Lastly, xy=;l-~==, which is a medial area (5}. 

Vi + X 

Thus [(3)1 (4), (5)] ( x +y) is ^ major irrational straight line. [x. 39] 

Actual solution gives 

and *+j--pVK i + \/7^ + 'V;( i- v7ta) 

The corresponding square root found in x. 94 is the minor irrational 
straight line, the terms being separated by a minus sign, and the two straight 
lines are the roots {cf. note on x. 5 1) of the biquadratic equation 

x* - lip 1 . x* + r iff* = o. 

r 1 + A r 



138 



BOOK X 



[x. 58 






Proposition 58. 



If an area be contained by a rational straight line and the 
fifth binomial, the "side " of the area is the irrational straight 
line called the side of a rational plus a medial area. 

For let ihe area AC be contained by the rational straight 
line AB and the fifth binomial AD divided into its terms at 
E, so that AE is the greater term ; 

I say that the "side" of the area AC is the irrational straight 
line called the side of a rational plus a medial area. 

For let the same construction be made as before shown ; 
it is then manifest that MO is the "side" of the area AC. 



Q E 



UK L C 



M 





1 Q 






O 


N 





It is then to be proved that MO is the side of a rational 
plus a medial area. 

For, since AG is incommensurable with GE, \%. 18] 

therefore AH is also commensurable with HE, [vi. 1, x. n] 
that is, the square on MN with the square on NO ; 
therefore MN, NO are incommensurable in square. 

And, since AD is a fifth binomial straight line, and ED 
the lesser segment, 
therefore ED is commensurable in length with AB. 

fx. Deff. 11. 5] 

But AE is incommensurable with ED ; 
therefore AB is also incommensurable in length with AE. 

[*• '3] 

Therefore AK, that is, the sum of the squares on MN, 
NO, is medial. [x. 21] 

And, since DE is commensurable in length with AB, that 
is. with EK, 

while DE is commensurable with EF, 
therefore EF is also commensurable with EK. [x. 1a) 



} ■ <■>■ 

; (ft 



x. 58] PROPOSITION 58 119 

And EK is rational ; 
therefore EL, that is, MR, that is, the rectangle MN, NO, is 
also rational. [x. 19] 

Therefore MN, NO are straight lines incommensurable 
in square which make the sum of the squares on them medial, 
but the rectangle contained by them rational. 

Therefore MO is the side of a rational plus a medial area 
[x. 40] and is the "side" of the area AC. 

Q. E. D. 

We have here to find the square root of the expression [cf. x. 52] 

p (hp J 1 + A + kp). 
As usual, we put 

u + v = kpji + A 

uv = i *y 

Then, u, v being found, we take 

x* = pu 
f = pv 
and {x +y), so fouuu, is our required square root. 

Euclid's proof of the class of (x +_>>) is as follows : 

By x. 18, « u v; 
therefore pu v pv, 
so that x 1 ~y, 
and x^~y (3). 

Next u + v -j kp 

"ft 

whence p (u + v), or (^ 4-y 1 ), is a medial area {4). 

Lastly, xy = \kf?, which is a rational area (5). 

Hence [(3), (4), (5)] {x+y) is the side of a rational plus a medial area. 

|x, 40] 
If we solve algebraically, we obtain 

* = ^U/rTA+,/A), 
and x +y = P y - (7i + A + ,A) + P V - U l + * ' 



2 " '-.*)" v a-T* v ' 

The corresponding " side " found in x. 95 is a straight lint which produces 

with a rational area a medial whole, being of the form [x -y), where x, y 

have the same values as above. 

The two square roots are (cf. note on x. 52) the roots of the biquadratic 

equation 

x l - 2 V 7 iTJi . x* + «y = o. 



«3« 



BOOK X 



E* 59 



Proposition 59. 

If an area be contained by a rational straight tine and the 
sixth binomial, the "side" of the area is the irrational straight 
line called the side of the sum of two medial areas. 

For let the area ABCD be contained by the rational 
straight line AB and the sixth binomial AD, divided into its 
terms at E, so that AE is the greater term ; 
I say that the "side" of AC is the side of the sum of two 
medial areas. 

Let the same construction be made as before shown. 

A Q E F D R Q 



H K 







N 





It is then manifest that MO is the "side" of AC, and 
that MN is incommensurable in square with NO. 

Now, since EA is incommensurable in length with AB, 
therefore EA, AB are rational straight lines commensurable 
in square only ; 

therefore AK, that is, the sum of the squares on MN, NO, 
is medial. [x. 21] 

Again, since ED is incommensurable in length with AB, 
therefore FE is also incommensurable with EK; [x. 13] 

therefore FE, EK are rational straight lines commensurable 
in square only ; 

therefore EL, that is, MR, that is, the rectangle MN, NO, is 
medial. [x. 21] 

And, since AE is incommensurable with EF, 
A A' is also incommensurable with EL. [vi. i, x. it] 

But AK is the sum of the squares on MN, NO, 
and EL is the rectangle MN, NO \ 

therefore the sum of the squares on MN, NO is incommen- 
surable with the rectangle MN. NO, 

And each of them is medial, and MN, NO are incom- 
mensurable in square. 



x. 59. Lemma] PROPOSITION 59 i 3 j 

Therefore MO is the side of the sum of two medial areas 
[x. 41], and is the " side " of AC. 

Q. K. D. 

Euclid here finds the square root of the expression [cf, x. 53] 

p(JA.p + J\. p). 
As usual, we solve the equations 

* + P = V*. P i 

p 5 J {h 



} (*). 



uv- ^kp 1 
then, », v being found, we put 

y = pv 
and {* + _y) is the square root required. 

Euclid proves that (x + y) is the side of {the sum of) two medial areas, as 
follows. 

As in the last two propositions, x, y are proved to be incommensurable 
in square. 

Now Jk , p, p are. commensurable in square only ; 
therefore p(u + v), or (x* +y 1 ), is a media/ area (3), 

Next, xy m % ,/A \ p\ which is again a media/ area (4). 

Lastly, Jfc . p « \ Jk . p, 
so that V*V ~i*/k.p*; 
'hat is, (x*+?) v xy „ (5). 

Hence [(3), (4), (5)] (x +_y) is the side of the sum of two medial areas. 

Solving the equations algebraically, we have 

and x+y = P J\ {Jk + <Jk=k) + p J^U* ~ J*^>- 

The corresponding square root found in x. 96 is x-y, where x,y are the 

same as here. 

The two square roots are (cf. note on x. 53) the roots of the biquadratic 

equation 

**-a./*.pV + (A-XV = o. 

t 

[Lemma. 

If a straight line be cut into unequal parts, the squares 
on the unequal parts are greater 

than twice the rectangle con- £ ? — £ P 

tained by the unequal parts. 

Let AB be a straight line, and let it be cut into unequal 
parts at C, and let A C be the greater ; 

I say that the squares on AC, CB are greater than twice the 
rectangle AC, CB. 









13* BOOK X [Lemma, x. 60 

For let AB be bisected at D. 

Since then a straight line has been cut into equal parts 
at D, and into unequal parts at C, 

therefore the rectangle AC, CB together with the square on 
CD is equal to the square on AD, [n. 5] 

so that the rectangle AC, CB is less than the square on AD; 
therefore twice the rectangle AC, CB is less than double of 
the square on AD. 

But the squares on AC, CB are double of the squares on 
AD, DC; [11. 9] 

therefore the squares on AC, CB are greater than twice the 
rectangle AC, CB. 

Q. e. a] 






We have already remarked {note on x. 44) that the Lemma here proving 
that 

x* + jp > 2xy 
can hardly be genuine, since the result is used in x. 44. 

Proposition 60. 

The square on the binomial straight line applied to a 
rational straight line produces as breadth the first binomial. 

Let AB be a binomial straight line divided into its terms 
at C, so that AC is the greater term ; 

let a rational straight line DE be ° £— y 

set out, 

and let DEFG equal to the square 

on AB be applied to DE producing 

DG as its breadth ; 

I say that DG is a first binomial 

straight line. 

For let there be applied to DE the rectangle DH equal 
to the square on AC, and KL equal to the square on BC; 
therefore the remainder, twice the rectangle AC, CB, is equal 
to MF. 

Let MG be bisected at A*", and let NO be drawn parallel 
[to ML or GF\ 

Therefore each of the rectangles MO, NF is equal to 
once the rectangle AC, CB. 

Now, since AB is a binomial divided into its terms at C, 



TTl 



x. 6o] PROPOSITION 60 133 

therefore AC, CB are rational straight lines commensurable 
in square only ; [x. 36] 

therefore the squares on AC, CB are rational and commen- 
surable with one another, 
so that the sum of the" squares or A C, CB is also rational. 

[x. , S ] 
And it is equal to DL ; 

therefore DL is rational. 

And it is applied to the rational straight line DE ; 
therefore DM is rational and commensurable in length with 
DE. [x. to] 

Again, since AC, CB are rational straight lines commen- 
surable in square only, 
therefore twice the rectangle AC, CB, that is ME, is medial. 

[X. 2l] 

And it is applied to the rational straight line ML ; 
therefore MG is also rational and incommensurable in length 
with ML, that is, DE. (x. 22] 

But MD is also rational and is commensurable in length 
with DE ; 
therefore DM is incommensurable in length with MG. [x. 13] 

And they are rational ; 

therefore DM, MG are rational straight lines commensurable 

in square only; 

therefore DG is binomial. [x. 36] 

It is next to be proved that it is also a first binomial 
straight line. 

Since the rectangle AC, CB is a mean proportional between 
the squares on AC, CB, [cf. Lemma after x. 53] 

therefore MO is also a mean proportional between DH, KL. 

Therefore, as DH is to MO, so is MO to KL, 

that is, as DK is to MN, so is MN to MK\ [vi. 1] 

therefore the rectangle DK, KM is equal to the square 
on MN. [vi. \<(\ 

And, since the square on AC is commensurable with the 
square on CB, 

DH is also commensurable with KL, 
so that DK is also commensurable with KM. [vi. 1, x. n] 



ij4 BOOK X [x. 60 

And, since the squares on AC, CB are greater than twice 
the rectangle AC, CB, [Lemma] 

therefore DL is also greater than MF, 
so that DM is also greater than MG, [vi. 1] 

And the rectangle DK, KM is equal to the square on 
MN, that is, to the fourth part of the square on MG, 
and DK is commensurable with KM. 

But, if there be two unequal straight lines, and to the greater 
there be applied a parallelogram equal to the fourth part of 
the square on the less and deficient by a square figure, and 
if it divide it into commensurable parts, the square on the 
greater is greater than the square on the less by the square 
on a straight line commensurable with the greater ; [x. 17] 
therefore the square on DM is greater than the square on 
MG by the square on a straight line commensurable with DM, 

And DM, MG are rational, 
and DM, which is the greater term, is commensurable in length 
with the rational straight line DE set out. 

Therefore DG is a first binomial straight line. [x. Deft", n. i] 

Q. E. D. 

In the hexad of propositions beginning with this we have the solution of 
the converse problem to that of X. 54 — 59. We find the squares of the 
irrational straight lines of x. 36 — 41 and prove that they are respectively equal 
to the rectangles contained by a rational straight line and the first, second, 
third, fourth, fifth and sixth binomials. 

In x. 60 we prove that, p + Jh . p being a binomial straight line [x. 36], 



is a first binomial straight line, and we find it geometrically. 
The procedure may b< 
Take x,y, t such that 



The procedure may be represented thus. 



ax = p*, 

try = kf? 
it . 2% = 2 Jk . p', 
p\ hp % being of course the squares on the terms of the original binomial, 
and 2 Jk.p 1 twice the rectangle contained by them. 

Then tf ^) t * r fe±^d£, 

and we have to prove that (x+y) + 21 is a. first binomial straight line of which 
(x+y), 2z are the terms and (x+y) the greater. 

Euclid divides the proof into two parts, showing first that {x + y) + 2$ is 
some binomial, and secondly that it is the first binomial. 



x. 6o, fil] PROPOSITIONS 60, 61 135 

(a) p 'w <J& . p, so that p 5 , ip* are rational and commensurable ; 

therefore p* + kp\ or cr (je +^), is a rational area, 

whence (x +y) is rational and " a (1). 

Next, 2p . Ji . p is a medial area, 
so that o- . 25 is a medial area, 
whence 2s is rational but ^ <r {2). 

Hence [(t), (2)] {-v+^}, iz are rational and commensurable in square 

only - (3); 

thus (x +y) + tz is a binomial straight line. [x. 36] 

W pO-.jk.p^jk.t?; kp\ 

SO that <r x ; trz = az : try, 

and x : z = z :y, 

or .xy^z'-liizf (4). 

Now p 1 , Ap* are commensurable, so that <r.v, try are coin mensurable, and 
therefore 

**y (s)- 

And, since [Lemma] p 3 + ip' > 2 JA . p a , 
x + y > 2Z. 

But (x + y) is given, being ei|ual to (6). 






Therefore [(4}, ($), (6), and x. 17] J(x +yy~(tjif « ( x +y). 
And (x +y), 2t are rational and "- [(3)], 
while (*+j») -«■[{!)]. 

Hence {x +y) + 2z is & first binomial. 
The actual value of (x +y) + 22 is, of course, 
pt> ___ 

- (l +-4 + 2 Jk). 

Proposition 61. 

The square on the first bimedial straight line applied to a 
rational straight line produces as breadth the second binomial. 

Let AB be a first bimedial straight line divided into its 
medials at C, of which medials AC 

is the greater; _ D K m n__g 

let a rational straight line DE be set 
out, 

and let there be applied to DE the 
parallelogram DF equal to the square 



on AB, producing DG as its breadth; e h l o f 

I say that DG is a second binomial a ~~6 8 

straight line. 

For let the same construction as before be made. 



136 BOOK X [x. 61 

Then, since AB is a first bimedial divided at C, 
therefore AC, CB are medial straight lines commensurable in 
square only, and containing a rational rectangle, [x. 37] 

so that the squares on AC, CB are also medial. [x, 21] 

Therefore DL is medial. [x, 15 and 23, Por.] 

And it has been applied to the rational straight line DE ; 
therefore MD is rational and incommensurable in length 
with DE. [x. 22] 

Again, since twice the rectangle A C, CB is rational, ME is 
also rational. 

And it is applied to the rational straight line ML ; 
therefore MG is also rational and commensurable in length 
with ML, that is, DE ; [x. 20] 

therefore DM is incommensurable in length with MG. [x. 13] 

And they are rational ; 
therefore DM, MG are rational straight lines commensurable 
in square only ; 
therefore DG is binomial. [x. 36] 

It is next to be proved that it is also a second binomial 
straight line. 

For, since the squares on AC, CB are greater than twice 
the rectangle AC, CB, 
therefore DL is also greater than ME, 
so that DM is also greater than MG. [vi. 1] 

And, since the square on AC is commensurable with the 
square on CB, 

DH is also commensurable with KL, 
so that DK is also commensurable with KM. [vi. 1, x. n] 

And the rectangle DK, KM is equal to the square on MN; 

therefore the square on DM is greater than the square on 

MG by the square on a straight line commensurable with DM, 

[x- 17] 
And MG is commensurable in length with DE, 

Therefore DG is a second binomial straight line. [x. Deff. 11. z\ 

In this case we have to prove that, ($ ? + $p) being a first bimedial 
straight tine, as found in X. 37, 

a 

is a stand binomial straight line. 



x. 6r, 6a] PROPOSITIONS 6i, 6* 137 

The form of the proposition ana the figure being similar to those of x. 60, 
I can somewhat abbreviate the reproduction of the proof. 
Take x, y, z such that 

ax = k 'p, 

a.xz = tip'. 
Then shall (x + y) + iz be a second binomial. 

(a) A*p, k'p.aie medial straight lines commensurable in square only and 
containing a rational rectangle. [x. 37} 

The squares £*p*> ^V are medial ; 
thus the sum, or o-{je +y), is medial. (x. 23, Por.] 

Therefore (x +y) is rational and ^ <r. 

And it . 2: is rational ; 
therefore 22 is rational and « <r (1). 

Therefore (£ + >■), as are rational and "- (2), 

so that (x +y) + 2z is a binomial. 

<fi) As before, (* +^) > as. 

Now, JPff, iPff being commensurable, 

And jry = s', 

... *v + *v 

while *+ v= — — . 

Hence [x. 17] J(x+y)' -^sf^ix+y) (3). 

But U « <r, by (1). 

Therefore [(1), (2), (3)] (jc +_>>) + a* is a second binomial straight line. 

Of course (x+y) + az = - {^(1 +/4) + a*}. 

IT 

Proposition 62. 

7"j4e square an the second bimedial straight line applied to 
a rational straight line produces as breadth the third binomial. 

Let AB be a second bimedial straight line divided into 
its medials at C, so that AC is the 
greater segment ; '- ' 

let DE be any rational straight line, 
and to DE let there be applied the 
parallelogram DF equal to the square 
on AB and producing DG as Its 
breadth ; 

I say that DG is a third binomial 
straight line. 

Let the same construction be made as before shown 



r H L O F 



i 3 8 BOOK X [x. 62 

Then, since AB is a second bimedial divided at C, 
therefore AC, CB are medial straight lines commensurahle in 
square only and containing a medial rectangle, [x, 38] 

so that the sum of the squares on AC, CB is also medial. 

[x. 15 and 23 Por.] 
And it is equal to DL ; 

therefore DL is also medial. 

And it is applied to the rational straight line DE ; 
therefore MD is also rational and incommensurable in length 
with DE. [x. zx\ 

For the same reason, 
MG is also rational and Incom mensurable in length with ML, 
that is, with DE ; 

therefore each of the straight lines DM, MG is rational and 
incommensurable in length with DE. 

And, since AC is incommensurable in length with CB, 
and, as AC is to CB, so is the square on AC to the rectangle 
AC, CB, 

therefore the square on A C is also incommensurable with the 
rectangle AC, CB. [x. n] 

Hence the sum of the squares on AC, CB is incommen- 
surable with twice the rectangle AC,' CB, [x. 12, 13] 
that is, DL is incommensurable with ME, 
so that DM is also incommensurable with MG. [vi. 1, x. 11] 

And they are rational ; 
therefore DG is binomial. [x. 36] 

It is to be proved that it is also a third binomial straight line. 

In manner similar to the foregoing we may conclude that 
DM is greater than MG, 
and that DK is commensurable with KM. 

And the rectangle DK, KM is equal to the square on 
MN; 

therefore the square on DM is greater than the square on 
MG by the square on a straight line commensurable with 
DM. 

And neither of the straight lines DM, MG is commen- 
surable in length with DE. 

Therefore DG is a third binomial straight line, [x, Deff. 11. 3] 

Q. E. D. 



x. 6z, 63] PROPOSITIONS 62, 63 139 

We have to prove that [cf. x, 38] 



1 /.i X W 



is a third binomial straight line. 
Take x, y, z such that 

<TX = A V, 

V 



^' 






CF . 2Z = 2 ,/A. . p>. 

{o) Now £*p, — ~ are medial straight lines commensurable in square only 

k* 
and containing a medial rectangle. [x. 38] 

The sum of the squares on them, or o- (* +y), is media/; 

therefore (x +y) is rational and ~ a (1). 

And ..: . 22 being medial also, 

2Z is rational and U a i 2 )- 

Now ^^(^^V-f 

= trx : trz, 
whence trx ~ o-z. 



But (A*pY «{^y + ra"}, or 



trjc a tr (jc + _y), and <tz « cr . 2a ; 



therefore <r(x +y) ~ a . tz, 

or {x+y)^> 2Z (3). 

Hence [{0, (2), (3)] (x +y) + 2z is a binomial straight line (4). 

(fi) As before, (x +y) > 2z, 

and *«^. 

Also xy = §*. 

Therefore [x. 17] •J(x+yf-(*z) 1 « (x+jr). 

And [(1), (2)] neither (jc + j') nor 22 is « tr. 

Therefore (* +.?) + 2Z is a Z^iVrf binomial straight line. 

Obviously (x +_y) + 2z ■ - j —jr + 2 V*[ ■ 

Proposition 63. 

The square on the major straight line applied to a rational 
straight tine produces as breadth the fourth binomial. 

Let AB be a major straight line divided at C, so that AC 
is greater than CB ; 
let DE be a rational straight line, 



D K M N Q 










E H U 



140 BOOK X [x. 63 

and to DE let there be applied the parallelogram DF equal 
to the square on AB and producing DG as Its breadth ; 
I say that DG is a fourth binomial 
straight line. 

Let the same construction be 
made as before shown. 

Then, since AB is a major 
straight line divided at C, 
AC, CB are straight lines incom- 
mensurable in square which make c e 
the sum of the squares on them 
rational, but the rectangle contained by them medial. [x. 39] 

Since then the sum of the squares on AC, CB is rational, 
therefore DL is rational ; 

therefore DM is also rational and commensurable in length 
with DE. [x. 20] 

Again, since twice the rectangle AC, CB, that is, ME, is 
medial, 

and it is applied to the rational straight line ML, 
therefore MG is also rational and incommensurable in length 
with DE ; [x. 22] 

therefore DM is also incommensurable in length with MG. 

Therefore DM, MG are rational straight lines commen- 
surable in square only ; 
therefore DG is binomial. [x. 36] 

It is to be proved that it is also a fourth binomial straight line. 

In manner similar to the foregoing we can prove that 
DM is greater than MG, 
and that the rectangle DK, KM is equal to the square on MN. 

Since then the square on AC is incommensurable with the 
square on CB, 

therefore DH is also incommensurable with KL, 
so that DK is also incommensurable with KM. [vi. 1, x. n] 

But, if there be two unequal straight lines, and to the 
greater there be applied a parallelogram equal to the fourth 
part of the square on the less and deficient by a square 
figure, and if it divide it into incommensurable parts, then the 



x. 6 3 ] PROPOSITION 63 141 

square on the greater will be greater than the square on the 
less by the square on a straight line incommensurable in 
length with the greater ; [x. 18] 

therefore the square on DM is greater than the square on 
MG by the square on a straight line incommensurable with 
DM. 

And DM, MG are rational straight lines commensurable 
in square only, 

and DM is commensurable with the rational straight line DE 
set out. 

Therefore DG is a fourth binomial straight line. [x. Deff. 11. 4] 

Q. E. D. 

We y v/e to prove that [cf, x. 39] 

is a fourth binomial straight line. 

For brevity we must call this expression 

-im% vf. 

tr v 
Take x, y, z such that 

<MC = «* 

try = 1? 

tr , 23 ZUV 

wherein it has to be remembered [x, 39] that u, v are incommensurable in 

square, {« ! + sr) is rational, and uv is medial. 

(a) («* + (/), and therefore a (x +y), is rational ; 

therefore {x +y) is rational and <■» a (1). 

3uv, and therefore a . zz, is medial ; 
therefore zz is rational and v a (2). 

Thus {■* +>), 2 * are rational and ^- (3), 

so that (x +y) + tz is a binomial straight line. 
{$) As before, x+y> iz, 

and #y = *"■ 

Now, since a' ^ tr 1 , 
ax u oj 1 , or x ^y. 

Hence [x. 18] *J(x +yy -(**)* ~(x +y) (4). 

And (x +y) « <r, by (1). 

Therefore [(3), (4)] (.* +.y) + 2* is a fourth binomial straight line. 

It is of course j» 1 1 + , 5 l . 



14* BOOK X [x. 64 



D 


CM H 










E H L 



Proposition 64. 

The square on the side of a rational plus a medial area 
applied to a rational straigltt line produces as breadth the fifth 
binomial. 

Let AB be the side of a rational plus a medial area, 
divided into its straight lines at C, 
so that AC is the greater ; 
let a rational straight line DE be set 
out, 

and let there be applied to DE the 
parallelogram DF equal to the square 
on AB, producing DG as its breadth ; 
I say that DG is a fifth binomial 
straight line. 

Let the same construction as before be made. 

Since then AB is the side of a rational plus a medial 
area, divided at C, 

therefore AC, CB are straight lines incommensurable in square 
which make the sum of the squares on them medial, but the 
rectangle contained by them rational. [x. 40] 

Since then the sum of the squares on AC, CB is medial, 
therefore DL is medial, 

so that DM is rational and incommensurable in length with 
DE. [x. »] 

Again, since twice the rectangle AC, CB, that is ME, is 
rational, 
therefore MG is rational and commensurable with DE. [x. 20] 

Therefore DM is incommensurable with MG ; [x. 13] 

therefore DM, MG are rational straight lines commensurable 
in square only ; 
therefore DG is binomial. [x. 36] 

I say next that it is also a fifth binomial straight line. 

For it can be proved similarly that the rectangle DK, KM 
is equal to the square on MN, 

and that DK is incommensurable in length with KM; 
therefore the square on DM is greater than the square on MG 
by the square on a straight line incommensurable with DM. 

[x. ,8] 



SL 



x. 64, 65] PROPOSITIONS 64, 65 143 

And DM, MG are commensurable in square only, and the 
less, MG, is commensurable in length with DE. 
Therefore DG is a fifth binomial. 

Q. E. D. 

To prove that [cf. x. 40] 

is ay^/A binomial straight line. 

1 
For brevity denote it by - (a + vf, and put 

(T . 2Z = 2«P. 

Remembering that [x. 40] n'^i?, («* + ii*) is medial, and 2uv is rational, 
we proceed thus. 

(o) o- (x ±y) is media] ; 

therefore (x +y) is rational and ur (1). 

Next, a . 2z is rational ; 
therefore as is rational and * o- (2). 

Thus (a - +,v), 2z are rational and <*- (3), 

so that (jr +J 1 ) + 22 is a binomial straight line, 
(J3) As before, * + y > 23, 

*? = **, 
and x uj 1 . 

Therefore [x. 18] %'(*+.)')* - ^J" „ (* +j<) (4). 

Hence [(2), (3), (4)] (x +y) + iz is a fifth binomial straight line. 

It is of course ■- \ , + „i . 






Proposition 65. 

The square on the side of the sum of two medial areas 
applied to a rational straight line produces as breadth the 
sixth binomial \ 

Let AB be the side of the sum of two medial areas, 
divided at C, 

let DE be a rational straight line, 

and let there be applied to DE the parallelogram DF equal 
to the Square on AB, producing DG as its breadth ; 



H4 



BOOK X 



L x. 65 



D K M N 










E H 






I say that DG is a sixth binomial straight line. 

For let the same construction be made as before. 

Then, since AB is the side of 
the sum of two medial areas, divided 
at C, 

therefore AC, CB are straight lines 
incommensurable in square which 
make the sum of the squares on 
them medial, the rectangle contained 
by them medial, and moreover the 
sum of the squares on them incom- 
mensurable with the rectangle contained by them, [x. 41] 

so that, in accordance with what was before proved, each of 
the rectangles DL, MF is medial. 

And they are applied to the rational straight line DE ; 

therefore each of the straight lines DM, MG is rational and 
incommensurable in length with DE. [x. *a] 

And, since the sum of the squares on AC, CB is incom- 
mensurable with twice the rectangle AC, CB, 

therefore DL is incommensurable with MF. 

Therefore DM is also incommensurable with MG ; 

[Vi, I, x. 11] 

therefore DM, MG are rational straight lines commensurable 
in square only ; 

therefore DG is binomial. [x. 36] 



I say next that it is also a sixth binomial straight line. 
Similarly again we can prove that the rectangle DK, KM 
is equal to the square on MN, 

and that DK is incommensurable in length with KM; 

and, for the same reason, the square on DM is greater than 
the square on MG by the square on a straight line incom- 
mensurable in length with DM. ■ 

And neither of the straight lines DM, MG is commen- 
surable in length with the rational straight line DE set out. 
Therefore DG is a sixth binomial straight line. 

Q. E. D. 



x. 6$, 66] PROPOSITIONS 65, 66 145 

To prove that [cf. x. 41] 

<r \J2 V Vl+<4 a -J* V <Jl + P> 

is a sixth binomial straight line. 

Denote it by - (u + v)', and put 

fT.3Z= 2UV. 

Now, by x. 41, «' v> tf, («* + ^) is medial, 2«jj is medial, and 

{«*+**) i/ 2«tf, 

(a) In this case a(x+y) is medial ; 

therefore {x +y) is rational and v a (t). 

In like manner, 22 is rational and - c (2). 

And, since <r (x +y) ^ <r . 2g, 

(x+y) o 23 (3). 

Therefore (x +y) + 22 is a binomial straight line. 
(jS) As before, x +y > 2z, 

xy = z*, 

therefore [x. 18] 7(*+>)*- (22)* *, (x + y) (4). 

Hence [(1), (2), (3), {4)] (x +y) + 2» is a rt'jfM binomial straight line. 

It is obviously - \ JK + -yS — j ■ 



Proposition 66. 

A straight line commensurable in length with a binomial 
straight line is itself also binomial and the same in order. 

Let AB be binomial, and let CD be commensurable in 
length with AB ; 

A 1 B 



I say that CD is binomial and the same in order with AB. 

For, since AB is binomial, 
let it be divided into its terms at B, 
and let AB be the greater term ; 



i 4 6 BOOK X [x. 66 

therefore AE, EB are rational straight lines commensurable 
in square only. [x. 36] 

Let it be contrived that, 
as AB is to CD, so is AE to CF\ [w. ia] 

therefore also the remainder EB is to the remainder FD as 
AB is to CD, [v. l9 ] 

But AB is commensurable in length with CD; 
therefore AE is also commensurable with CF, and EB with 
FD. [x. 11] 

And ^Zj, EB are rational ; 
therefore CF, FD are also rational. 

And, as AE is to CF, so is /TZ? to FD. [v. 11] 

Therefore, alternately, as AE is to ,£7?, so is CF to Z 7 "/?. 

[v. 16] 

But ^4-ZT, ZTZ? are commensurable in square only ; 

therefore CF t FD are also commensurable in square only. 

[x. .I] 
And they are rational ; 

therefore CD is binomial. [x. 36] 

I say next that it is the same in order with AB. 

For the square on AE is greater than the square on EB 
either by the square on a straight line commensurable with 
AE or by the square on a straight line incommensurable 
with it. 

If then the square on AE is greater than the square on 
EB by the square on a straight line commensurable with AE, 
the square on CF will also be greater than the square on FD 
by the square on a straight line commensurable with CF. 

[x. '4] 

And, if AE is commensurable with the rational straight 
line set out, CF will also be commensurable with it, [x. 12] 

and for this reason each of the straight lines AB, CD is a 

first binomial, that is, the same in order. [x. Deff - "■ l ] 

But, if EB is commensurable with the rational straight line 

set out, FD is also commensurable with it, [x. i*] 

and for this reason again CD will be the same in order with 
AB, 

for each of them will be a second binomial. [x, Deff. n. *] 



x. 66, 67] PROPOSITIONS 66, 67 147 

But, if neither of the straight lines AE, EB is commen- 
surable with the rational straight line set out, neither of the 
straight lines CF, FD will be commensurable with it, [x. 13] 

and each of the straight lines AB, CD is a third binomial, 

[x. Deff. 11. 3] 
But, if the square on AE is greater than the square on 
EB by the square on a straight line incommensurable with 
AE, 

the square on CF is also greater than the square on FD by 
the square on a straight line incommensurable with CF. [x. 14] 

And, if AE is commensurable with the rational straight 
line set out, CF is also commensurable with it, 
and each of the straight lines AB, CD is a fourth binomial. 

[x. Deff. 11. 4] 

But, if EB is so commensurable, so is FD also, 

and each of the straight lines AB, CD will be a fifth binomial. 

[x. Deff. 11. 5] 
But, if neither of the straight lines AE, EB is so com- 
mensurable, neither of the straight lines CF, FD is commen- 
surable with the rational straight line set out, 

and each of the straight lines AB, CD will be a sixth binomial. 

[x. Deff. it, 6] 
Hence a straight line commensurable in length with a 
binomial straight line is binomial and the same in order. 

Q. e, o. 

The proofs of this and the following propositions up to x. 70 inclusive ate 
easy and require no elucidation. They are equivalent to saying that, if in each 

of the preceding irrational straight lines — p is substituted for p, the resulting 

irrational is of the same kind as that from which it is altered. 

Proposition 67. 

A straight line commensurable in length with a hi me dial 
straight line is itself also bimedial and the same in order. 

Let AB be bimedial, and let CD be commensurable in 
length with AB; 

I say that CD is bimedial and the same *_ ? e 

in order with AB. c £_p 

For, since AB is bimedial, 
let it be divided into its mediate at E ; 



i4» BOOK X [x. 67 

therefore AE, EB are medial straight lines commensurable 
in square only. [x. 37, 38] 

And let it be contrived that, 

as AB is to CD, so is AE to CF; 

therefore also the remainder EB is to the remainder FD as 
AB is to CD. [v. 19] 

But AB is commensurable in length with CD ; 

therefore AE, EB are also commensurable with CF t FD 

respectively. [x. 11] 

But AE, EB are medial ; 
therefore CF, FD are also medial. [x. 23] 

And since, as A E is to EB, so is CF to FD, [v. 1 r] 

and AE, EB are commensurable in square only, 
CF, FD are also commensurable in square only. [x. n] 

But they were also proved medial ; 
therefore CD is bimedial. 

I say next that it is also the same in order with AB. 
For since, as AE is to EB, so is CF to FD, 

therefore also, as the square on AE is to the rectangle AE, 
EB, so is the square on CF to the rectangle CF, FD ; 

therefore, alternately, 

as the square on AE is to the square on CF, so is the rect- 
angle AE, EB to the rectangle CF, FD. [v. 16] 
But the square on AE is commensurable with the square 
on CF; 

therefore the rectangle AE, EB is also commensurable with 
the rectangle CF, FD, 

If therefore the rectangle AE, EB is rational, 

the rectangle CF, FD is also rational, 

[and for this reason CD is a first bimedial] ; [x. 37] 

but if medial, medial, [x. *3> P°r] 

and each of the straight lines AB, CD is a second bimedial. 

[X.J8] 

And for this reason CD will be the same in order with AB. 

Q. E. D. 



x. 68] PROPOSITIONS 67, 68 149 

Proposition 68. 

A straight line commensurable with a major straight 
line is itself also major. 

Let AB be major, and let CD be commensurable with AB; 
I say that CD is major. 

Let AB be divided at E ; 
therefore AE, EB are straight lines incommensur- 
able in square which make the sum of the squares 
on them rational, but the rectangle contained by 
them medial. [x. 39] 

Let the same construction be made as before. 

Then since, as AB is to CD, so is AE to CF, and EB 
to FD, 
therefore also, as AE is to CF, so is EB to FD. [v. 11] 

But AB is commensurable with CD; 
therefore AE, EB are also commensurable with CF, FD 
respectively, [x. n] 

And since, as AE is to CF, so is EB to FD, 
alternately also, 

as AE is to EB, so is CF to FD \ [v, 16] 

therefore also, componendo, 

as AB is to BE, so is CD to DF; [v. 18] 

therefore also, as the square on AB is to the square on BE, 
so is the square on CD to the square on DF. [vi. 20] 

Similarly we can prove that, as the square on AB is to 
the square on AE, so also is the square on CD to the square 
on CF. 

Therefore also, as the square on AB is to the squares on 
AE, EB, so is the square on CD to the squares on CF. FD ; 
therefore also, alternately, 

as the square on AB is to the square on CD, so are the 
squares on AE, EB to the squares on CF, FD. [v. 16] 

But the square on AB is commensurable with the square 
on CD ; 

therefore the squares on AE, EB are also commensurable 
with the squares on CF, FD. 



150 BOOK X [x. 68, 69 

And the squares on AE, EB together are rational ; 
therefore the squares on CF, FD together are rational, 

* Similarly also twice the rectangle AE, EB is commen- 
surable with twice the rectangle CF, FD. 

And twice the rectangle AE, EB is medial ; 
therefore twice the rectangle CF, FD is also medial. 

[x. *3, Por.] 

Therefore CF, FD are straight lines incommensurable in 
square which make, at the same time, the sum of the squares 
on them rational, but the rectangle contained by them medial; 
therefore the whole CD is the irrational straight line called 
major. [x. 39] 

Therefore a straight line commensurable with the major 
straight line is major. 

Q. E. D. 

Proposition 69. 

A straight line commensurable with the side of a rational 
plus a medial area is itself also the side of a rational plus a 
medial area. 

Let AB be the side of a rational plus a medial area, 
and let CD be commensurable with AB; 
it is to be proved that CD is also the side of a 
rational plus a medial area. 

Let AB be divided into its straight lines at E\ 
therefore AE, EB are straight lines incommensur- 
able in square which make the sum of the squares 
on them medial, but the rectangle contained by them 
rational. [x. 40] 

Let the same construction be made as before. 

We can then prove similarly that 
CF, FD are incommensurable in square, 
and the sum of the squares on AE, EB is commensurable 
with the sum of the squares on CF, FD, 
and the rectangle AE, EB with the rectangle CF, FD ; 
so that the sum of the squares on CF, FD is also medial, and 
the rectangle CF, FD rational. 

Therefore CD is the side of a rational plus a medial area. 

Q. E. D. 



a 



x. 7<>, ?i] PROPOSITIONS 68—71 151 



F 



Proposition 70. 

^4 straight line commensurable with the side of the sum 
of two medial areas is the side of the sum of two medial areas. 

Let AB be the side of the sum of two medial areas, and 
CD commensurable with AB \ 

it is to be proved that CD is also the side of the 
sum of two medial areas. 

For, since AB is the side of the sum of two 
medial areas, 

let it be divided into its straight lines at E ; 
therefore AE, EB are straight lines incommensur- 
able in square which make the sum of the squares 
on them medial, the rectangle contained by them 
medial, and furthermore the sum of the squares on AE, EB 
incommensurable with the rectangle AE, EB. [x. 41] 

Let the same construction be made as before. 

We can then prove similarly that 
CE, ED are also incommensurable in square, 
the sum of the squares on AE, EB is commensurable with 
the sum of the squares on CE, FD, 
and the rectangle AE, EB with the rectangle CF, FD \ 
so that the sum of the squares on CF, FD is also medial, 
the rectangle CF, FD is medial, 

and moreover the sum of the squares on CF, FD is incom- 
mensurable with the rectangle CF, FD. 

Therefore CD is the side of the sum of two medial areas. 

Q. E. D. 

Proposition 71. 

If a rational and a medial area be added together, four 
irrational straight lines arise, namely a binomial or a first 
bimedial or a major or a side of a rational plus a medial 
area, 

het.AB be rational, and CD medial ; 
I say that the "side" of the area AD is a binomial or a first 
bimedial or a major or a side of a rational plus a medial 
area. 



152 



BOOK X 



[x. 71 



For IB is either greater or less than CD, 

First, let it be greater ; 
let a rational straight line EF be set out, 
let there be applied to EF the rectangle EG equal to AB, 
producing EH as breadth, 

and let HI, equal to DC, be applied to EF, producing HK 
as breadth. 

A c 










F Q 



Then, since AB is rational and is equal to EG, 
therefore EG is also rational. 

And it has been applied to EF, producing EH as breadth; 
therefore EH is rational and commensurable in length with 
EF. [x. 20] 

Again, since CD is medial and is equal to HI, 
therefore HI is also medial. 

And it is applied to the rational straight line EF, pro- 
ducing HK as breadth ; 

therefore HK is rational and incommensurable in length 
with EF [x. 22] 

And, since CD is medial, 
while AB is rational, 

therefore AB is incommensurable with CD, 
so that EG is also incommensurable with HI. 

But, as EG is to HI, so is EH to HK; [vi. 1] 

therefore EH is also incommensurable in length with HK. 

[x. 11] 

And both are rational ; 

therefore EH, HK are rational straight lines commensurable 

in square only ; 

therefore EK is a binomial straight line, divided at H. [x. 36] 



x. 7«] PROPOSITION 71 153 

And, since AB is greater than CD, 
while AB is equal to EG and CD to HI, 
therefore EG is also greater than HI ; 
therefore EH is also greater than HK. 

The square, then, on EH is greater than the square on 
HK either by the square on a straight line commensurable 
in length with EH or by the square on a straight line in- 
commensurable with it. 

First, let the square on it be greater by the square on a 
straight line commensurable with itself. 

Now the greater straight line HE is commensurable in 
length with the rational straight line EF set out ; 
therefore EK is a first binomial. [x. Deff. 11. t] 

But EF is rational ; 
and, if an area be contained by a rational straight line and the 
first binomial, the side of the square equal to the area is 
binomial. [x. 34] 

Therefore the "side" of EI is binomial ; 
so that the "side" of AD is also binomial. 

Next, let the square on EH be greater than the square 
on HK by the square on a straight line incommensurable 
with EH. 

Now the greater straight line EH is commensurable in 
length with the rational straight line EF set out ; 
therefore EK is a fourth binomial. [x. Deff. 11. 4] 

But EF is rational ; 
and, if an area be contained by a rational straight line and the 
fourth binomial, the "side" of the area is the irrational straight 
line called major. [x. 57] 

Therefore the "side" of the area EI is major ; 
so that the "side" of the area AD is also major. 

Next, let AB be less than CD ; 
therefore EG is also less than HI 
so that EH is also less than HK, 

Now the square on HKis greater than the square on EH 
either by the square on a straight line commensurable with 
HK or by the square on a straight line incommensurable 
with it. 



154 BOOK X [x. 7i 

First, let the square on it be greater by the square on a 
straight line commensurable in length with itself. 

Now the lesser straight line EH is commensurable in 
length with the rational straight line EF set out ; 
therefore EK is a second binomial. [x. Deff. 11. 2] 

But EF is rational , 
and, if an area be contained by a rational straight line and 
the second binomial, the side of the square equal to it is a 
first bimedial ; \x, 55] 

therefore the "side" of the area EI is a first bimedial, 
so that the "side" of AD Is also a first bimedial. 

Next, let the square on HK be greater than the square 
on HE by the square on a straight line incommensurable 
with HK. 

Now the lesser straight line EH is commensurable with 
the rational straight line EF set out ; 
therefore EK is a fifth binomial. [x. DeflT. u. 5] 

But EF is rational ; 
and, if an area be contained by a rational straight line and the 
fifth binomial, the side of the square equal to the area is a 
side of a rational plus a medial area. [x, 58} 

Therefore the "side" of the area EI is a side of a rational 
plus a medial area, 

so that the "side" of the area AD is also a side of a rational 
plus a medial area. 

Therefore etc. Q. E. d. 

A rationed area being of the form ip\ and a medial area of the form 
J\ . p", the problem is to classify 

according to the different possible relations between k, A. 
Put <rw = Ap*, 

av = JK. p\ 
Then, since the former rectangle is rational, the latter medial, 
u is rational and « a, 
v is rational and 'v <r. 
Also the rectangles are incommensurable ; 
so that u v v. 

Hence a, v are rational and «-j 
whence (u + v) is a binomial straight line. 



x. 7i, 7*] 



PROPOSITIONS 71, 7* 



*55 






The possibilities now are as follows : 

I. u > v. 
Then either 
{1) -Jif-tf " «, 

or (z) J if -^V ^ u, 
while in both cases w « a-. 

In case (1) (a + u) is a jf«/ binomial straight line, 
and in case (2} {« + f>) is a fourth binomial straight line. 

Thys -Jit(u + v) is either (1) a binomial straight line [x. 54] or (z) a major 
irrational straight line [x. 57]. 

II. p > «. 
Then either 






(i) -Jv* - u 1 « p, 
or (a) •Jtf — ti' « i/, 
while in both cases » ^ ir, but a " o-. 

Hence, in case (1), (v + it) is a JAVnif binomial straight line, 
and, in case {2}, (v + u) is & fifth binomial straight line. 

Thus V*t{t' + u) is either (1) a first bimedial straight line [x. 55], or (2} a 
side of a rational plus a medial area [x. 58]. 



Proposition 72. 

If two medial areas incommensurable with one another be 
added together, the remaining two irrational straight lines 
arise, namely either a second bimedial or a side of the sum 0/ 
two medial areas. 

For let two medial areas AB, CD incom mensurable with 
one another be added together ; 

I say that the "side" of the area AD is either a second 
bimedial or a side of the sum of two medial areas. 




For AB is either greater or less than CD, 
First, if it so chance, let AB be greater than CD. 
Let the rational straight line EEbe set out, 
and to EF let there be applied the rectangle EG equal to 



156 BOOK X [x. ft 

AB and producing EH as breadth, and the rectangle HI 
equal to CD and producing HK as breadth. 

Now, since each of the areas AB, CD is medial, 

therefore each of the areas EG, HI is also medial. 

And they are applied to the rational straight line FE, 
producing EH, HK as breadth ; 

therefore each of the straight lines EH, HK is rational and 
incommensurable in length with EF. [x. n] 

And, since AB is incommensurable with CD, 
and AB is equal to EG, and CD to HI, 
therefore EG is also incommensurable with HI. 

But, as EG is to HI, so is EH to HK ; [vi. i] 

therefore EH fa incommensurable in length with HK [x. it] 

Therefore EH, HK are rational straight lines commen- 
surable in square only ; 
therefore EK is binomial. [x. 3G] 

But the square on EH is greater than the square on HK 
either by the square on a straight line commensurable with 
EH or by the square on a straight line incommensurable 
with it. 

First, let the square on it be greater by the square on a 
straight line commensurable in length with itself. 

Now neither of the straight lines EH, II K is commen- 
surable in length with the rational straight line EF set out ; 
therefore EK is a third binomial. [x. Deff, n, 3] 

But EF is rational ; 
and, if an area be contained by a rational straight line and the 
third binomial, the "side" of the area is a second bi medial ; 

[x-5*] 

therefore the "side" of EI, that is, of AD, is a second bimedial. 

Next, let the square on EH be greater than the square 
on HK by the square on a straight line incommensurable in 
length with EH. 

Now each of the straight lines EH, HK is incommen- 
surable in length with EF; 

therefore EK is a sixth binomial. [x. Deff. 11. 6] 

But, if an area be contained by a rational straight line and 



x. 7*] PROPOSITION 72 157 

the sixth binomial, the "side" of the area is the side of the 
sum of two medial areas; [x. 59] 

so that the "side" of the area AD is also the side of the 
sum of two medial areas. 

Therefore etc. 

Q. E. D. 

We have to classify, according to the different possible relations between 
k, A, the straight line 

where *jk . p 1 and ,/A ■ p 5 are incommensurable. 

Suppose that au = Jk . p*, 

trv = Jk . p : . 

It is immaterial whether ^Jk.p 1 or ,/A. p- is the greater. Suppose, e.g., 
that the former is. 

Now, J 'A . p\ Jk . p' being both media/ areas, and a rational, 

u, ware both rational and ^ tr (1). 

Again, by hypothesis, uu ^ of, 

or u u v (2). 

Hence [(1), (»)] (u + v) is a binomial straight line. 

Next, Ju' - 1? is either commensurable or incommensurable in length 
with «. 

(a) Suppose v «* - ? rt «. 

In this case (« + t>) is a /A«>rf binomial straight line, 
and therefore [x, 56] 

«/»(» + ») is a second bitntdial straight line. 

03) If -Jtf^i? ii a, 

(« + ?') is a sf'.vrf binomial straight line, 

and therefore [x. 59] 

J a (k ■+• v) is a «ife 0/" /A; jkot fir" ftfff medial areas. 



The binomial straight line and the irrational straight lines 
after it are neither the same with the medial nor with one 
another. 

For the square on a medial, if applied to a rational straight 
line, produces as breadth a straight line rational and incom- 
mensurable in length with that to which it is applied. [x. 22] 

But the square on the binomial, if applied to a rational 
straight line, produces as breadth the first binomial. [x. 60] 

The square on the first bimedial, if applied to a rational 
straight line, produces as breadth the second binomial, [x. 61] 



BOOK X [x. 72, 73 

The square on the second bimedial, if applied to a rational 
straight line, produces as breadth the third binomial, [x. 61] 

The square on the major, if applied to a rational straight 
line, produces as breadth the fourth binomial, [x. 63] 

The square on the side of a rational plus a medial area, if 
applied to a rational straight line, produces as breadth the fifth 
binomial. [x. 64] 

The square on the side of the sum of two medial areas, if 
applied to a rational straight line, produces as breadth the 
sixth binomial. [x. 65] 

And the said breadths differ both from the first and from 
one another : from the first because it is rational, and from 
one another because they are not the same in order ; 
so that the irrational straight lines themselves also differ from 
one another. 

The explanation after x. 72 is for the purpose of showing that all the 
irrational straight lines treated hitherto are different from one another, viz. the 
medial, the six irrational straight lines beginning with the binomial, and the 
six consisting of the first, second, third, fourth, fifth and sixth binomials. 

Proposition 73, 

If from a rational straight line there be subtracted a 
rational straight line commensurable with the whole in square 
only, the remainder is irrational; and let it be called an 
apotome. 

For from the rational straight line AB let the rational 
straight line BC, commensurable with 
the whole in square only, be sub- a c b 

traded ; 

I say that the remainder AC is the irrational straight line 
called apotome. 

For, since AB is incommensurable in length with BC, 

and, as AB is to BC, so is the square on AB to the rectangle 
AB, BC, 

therefore the square on AB is incommensurable with the 
rectangle AB, BC jx. 11] 

But the squares on AB, BC are commensurable with the 
square on AB, [x. 15] 

and twice the rectangle AB, BC is commensurable with the 
rectangle AB, BC. [x. 6] 



x. 73. 74] PROPOSITIONS 72—74 159 

And, inasmuch as the squares on AB, BC are equal to 
twice the rectangle AB, BC together with the square on CA, 

[«■ 7] 
therefore the squares on AB, BC are also incommensurable 
with the remainder, the square on AC. [x. 13, 16] 

But the squares on AB, BC are rational ; 
therefore AC is irrational. [x. Def. 4] 

And let it be called an apotome. 

Q. E. D. 

Euclid now passes to the irrational straight lines which are the difference 
and not, as before, the sum of two straight lines. Apotome {"portion cut off") 
accordingly takes the place of binomial and the other terms follow mutatis 
mutandis. The first hcxad of propositions (73 to 78) exhibit the six irrational 
straight lines which are really the result of extracting the st/uare root of the six 
irrationals in the later propositions 85 to 90 {or, strictly speaking, of finding 
the sides of squares equal to the rectangles formed by each of those six 
irrational straight lines respectively with a rational straight tine). Thus, just 
as in the corresponding propositions about the irrational straight lines formed 
by addition, the further removed irrationals, so to speak, come first. 

We shall denote the apotome etc. by (x -y), which is formed by subtracting 
a certain lesser straight line y from a greater x. In x. 79 and later propositions 
y is called by Euclid the annex (>J B-rxwa/j^o^ouo-o}, being the straight line which, 
when added to the apotome or other irrational formed by subtraction, makes 
up the greater x. 

The methods of proof are exactly the same as in the preceding propositions 
about the irrational straight lines formed by addition. 

In this proposition x, y are rational straight lines commensurable in square 
only, and we have to prove that (x -y), the apotome, is irrational. 

x «<■ y , so that x v y : 
therefore, since x -.y-x 1 : xy, 

x* \j xy. 

But x 1 « (xr +yr), and xy *> txy ; 
therefore x* + y' ~ sxy, 

whence (x -y) ! ~ (x' +y*). 

But {x? +y) is rational , 
therefore (x ~y)', and consequently (x -y), is irrational. 

The apotome (x —y) is of the form p~ s /i . p, just as the binomial straight 
line is of the form p+ JA.p, 

Proposition 74. 

If from a medial straight tine there be subtracted a medial 
straight line which is commensurable with the whole in square 
only, and which contains with the whole a rational rectangle, 
the remainder is irrational. And let it be called a first 
apotome of a medial straight line. 






160 BOOK X [x. 74 

For from the medial straight line AB let there be sub- 
tracted the medial straight line BC 

which is commensurable with AB in A c ? 

square only and with AB makes the 
rectangle AB, BC rational ; 

I say that the remainder AC is irrational; and let it be 
called a first apotome of a medial straight line. 

For, since AB, BC are medial, 

the squares on AB, BC are also medial. 

But twice the rectangle AB, BC is rational ; 

therefore the squares on AB, BC are incommensurable with 
twice the rectangle AB, BC ; 

therefore twice the rectangle AB, BC is also incommensurable 
with the remainder, the square on AC, [cf. h. 7] 

since, if the whole is incommensurable with one of the magni- 
tudes, the original magnitudes will also be incommensurable. 

[x. 16] 
But twice the rectangle AB, BC is rational ; 

therefore the square on AC is irrational ; 

therefore AC is irrational. [x. Def. 4] 

And let it be called a first apotome of a medial straight 
line. 

TiiR first apotome of a medial straight line is the difference between straight 

lines of the form Hrp, Irp, which are medial straight lines commensurable in 
square only and forming a rational rectangle. 

By hypothesis, x 1 , y* are medial areas. 

And, since xy is rational, {x 1 +y) v xy 

~2xy, 
whence (*—yY ** **V' 

But 2xy is rational ; 
therefore (x — y)', and consequently (x —y), is irrational. 

This irrational, which is of the form (A*p ~ k'p), is the first apotome of a 
medial straight line ; the term corresponding of course to first bimedial, which 
applies where the sign is positive. 






75] 



PROPOSITIONS 74, 7S 



t6i 






Proposition 75. 



If from a medial straight line there be subtracted a medial 
straight line which is commensurable with the whole in square 
only, and which contains with the whole a medial rectangle, 
the remainder is irrational ; and let it be called a second 
apotomc of a medial straight line. 

For from the medial straight line AB let there be sub- 
tracted the medial straight line CB which is commensurable 
with the whole AB in square only and such that the rectangle 
AB, BC, which it contains with the whole AB, is medial; [x. 28] 

I say that the remainder AC is irrational ; and let It be called 
a second apotome of a medial straight line. 



a c 



FQ 







For let a rational straight line DI be set out, 
fet DE equal to the squares on AB, BC be applied to DI, 
producing DG as breadth, 

and let DH equal to twice the rectangle AB, BC be applied 
to DI, producing DF as breadth ; 

therefore the remainder FE is equal to the square on AC. 

["• 7] 

Now, since the squares on AB, BC are medial and 
commensurable, 
therefore DE is also medial, [x. 15 and 13, Por.] 

And it is applied to the rational straight line DI, producing 
DG as breadth ; 

therefore DG is rational and incommensurable in length 
with DI. [x. **] 

Again, since the rectangle AB, BC is medial, 

therefore twice the rectangle AB, BC is also medial. 

[x. »3, Por.] 



i6s BOOK X [x. 75 

And it is equal to DH ; 
therefore DH is also medial. 

And it has been applied to the rational straight line DI, 
producing DF as breadth ; 

therefore DF is rational and incommensurable in length 
with DI, [x. i*] 

And, since AB, BC are commensurable in square only, 
therefore AB is incommensurable in length with BC; 
therefore the square on AB is also incommensurable with the 
rectangle AB, BC. [x. n] 

But the squares on AB, BC are commensurable with the 
square on AB, [x. 15] 

and twice the rectangle AB, BC is commensurable with the 
rectangle AB, BC ; [x. 6] 

therefore twice the rectangle AB, BC is incommensurable with 
the squares on AB, BC. |x- 13] 

But DE is equal to the squares on AB, BC, 
and DH to twice the rectangle AB, BC; 
therefore DE is incommensurable with DH. 

But, as DE is to DH, so is GD to DF; [vi. i] 

therefore GD is incommensurable with DF. [x. 11] 

And both are rational ; 
therefore GD, DFare rational straight lines commensurable 
in square only ; 
therefore FG is an apotome. [x. 73] 

But DI is rational, 
and the rectangle contained by a rational and an irrational 
Straight line is irrational, [deduction from x, 20] 

and its "side" is irrational. 

And AC is the "side" of FE ; 
therefore AC is irrational. 

And let it be called a second apotome of a medial 
straight line. 

Q. E. D. 

We have here the difference between $p, J\.pj$, two medial straight 
lines commensurable in square only and containing a medial rectangle. 

Apply each of the areas (x'+j?), zxy to a rational straight line tr, i.e. 
suppose that 

x , +y = <ru, 

2xy = av. 



x. 75. 76] PROPOSITIONS 75. 7* 163 

Then <rw, <rt> are medial areas, 
so that u, v are both rational and « <r ,(1). 

Again, x v y ; 
therefore «* « xy, 
and consequently x* +>" ~ a*y, 
or rav <ro, 
and awo (i). 

Thus ((1), (*)] », vare rational and «-; 
therefore [x. 73] (« - v) is an apotome, 
and, (« — 1>) being thus irrational, 

(u — v) cr is an irrational area. 

Hence (x -y) 1 , and consequently (^ -y), is irrational. 

The irrational straight line A*p ~ ** -- ^ - is called a second apotome of a 
medial straight line. 



Proposition 76. 

If from, a straight line there be subtracted a straight line 
which is incommensurable in square with the whole and which 
with the whole makes the squares on them added together 
rational, but the rectangle contained by them medial, the 
remainder is irrational; and let it be called minor. 

For from the straight line AB let there be subtracted the 
straight line BC which is incom- 
mensurable in square with the whole a 4~ b 
and fulfils the given conditions. [x. 33] 

I say that the remainder A C is the irrational straight line 
called minor. 

For, since the sum of the squares on AB, BC is rational, 
while twice the rectangle AB, BC is medial, 
therefore the squares on AB, BC are incommensurable with 
twice the rectangle AB, BC; 

and, convertendo, the squares on AB, i?Care incommensurable 
with the remainder, the square on AC [11. 7, x. 16] 

But the squares on AB, BC are rational ; 
therefore the square on A C is irrational ; 
therefore AC is irrational. 

And let it be called minor. 






i6 4 BOOK X [x. 76, 77 

x, y are here of the form found in x. 33, viz. 

P / T p / J~ 

J*V t I + JTTF T^V 1 JTTe' 

By hypothesis (a 3 + y*) is a rational, xy a medial, area. 
Therefore {jp*+y) ^ 2Ay, 

whence (a -ji)' ^ (a 3 +^1- 

Therefore (a- -y)', and consequently (a — /), is irrational 
The minor (irrational) straight line is thus of the form 

Observe the use of tonvtrttndo (acnirTptyam) for the inference that, since 
(a* +y) « 2aji, (a* +_>>■) u (a -_v) s . The use of the word corresponds exactly 
to its use in proportions. 

Proposition 77. 

If from a straight line there be subtracted a straight line 
which is incommensurable in square with the whole, and which 
with the whole makes the sum of the squares on them medial, 
but twice the rectangle contained by them rational, the remainder 
is irrational: and let it be called that which produces with 
a rational area a medial whole. 

For from the straight line AB let there be subtracted the 
straight line BC which is incommensurable in square 
with AB and fulfils the given conditions ; [x. 34] 

I say that the remainder AC is the irrational straight 
line aforesaid. 

For, since the sum of the squares on AB, BC is 
medial, 

while twice the rectangle AB, BC is rational, 

therefore the squares on AB, BC are incommensurable 
with twice the rectangle AB, BC ; 

therefore the remainder also, the square on AC, is incom- 
mensurable with twice the rectangle AB, BC. [11. 7, x. 16] 

And twice the rectangle AB, BC is rational j 
therefore the square on AC is irrational ; 
therefore AC is irrational. 

And let it be called that which produces with a 
rational area a medial whole. 

Q. E. D. 



x. 77. 7»] PROPOSITIONS 76—78 165 

Here x, y are of the form [cf. x. 34] 



JjjTP + i, 









■J2(t+P) ' -Ja(i+JP) 

By hypothesis, (x 1 +y*) is a medial, xy a rational, area ; 
thus (^+y) m 2xc, 

and therefore {■*->)* m My, 

whence (x -yf, and consequently {x -y), is irrational. 
The irrational straight line 



is called thai which products with a rational area a medial whole or more 
literally that which with a rational area makes the whole medial {jJ ft«ri pijicX 
p. wav to $\ov troiovwn). Here " produces " means " produces when a square 
is described on it" A clearer way of expressing the meaning would be to call 
this straight tine the " side " of a medial minus a rational area corresponding 
to the "side" of a rational plus a medial area [x. 40]. 



Proposition 78. 

If from a straight line there be subtracted a straight line 
■which is incommensurable in square with the whole and which 
Tvith the whole makes the sum of the squares on them medial, 
twice the rectangle contained by them medial, and further the 
squares on them incommensurable with twice the rectangle 
contained by them, the remainder is irrational ; and let it be 
called that which produces with a medial area a 
medial whole. 

For from the straight line AB let there be subtracted the 
straight line BC incommensurable in 

square with AB and fulfilling the p f q 

given conditions ; [x. 35] 

I say that the remainder AC is the 
irrational straight line called that 
which produces with a medial 
area a medial whole. 

For let a rational straight line DI * c a 

be set out, 

to DI let there be applied DE equal to the squares on AB, 
BC, producing DG as breadth, 

and let DH equal to twice the rectangle AB, BC be 
subtracted. 




i« BOOK X [x. 78 

Therefore the remainder FE is equal to the square 
on AC, [u. 7] 

so that AC is the " side " of FE. 

Now, since the sum of the squares on AB, BC is medial 
and is equal to DE, 

therefore DE is medial. 

And it is applied to the rational straight line DI, producing 
DG as breadth ; 

therefore DG is rational and incommensurable in length 
with DI. [x. *»] 

Again, since twice the rectangle AB, BC is medial and is 
equal to DH, 

therefore DH is medial. 

And it is applied to the rational straight line DI, producing 
DF as breadth ; 

therefore DF is also rational and incommensurable in length 
with DI. [x. 12] 

And, since the squares on AB, BC are incommensurable 
with twice the rectangle AB, BC, 

therefore DE is also incommensurable with DH. 

But, as DE is to DH, so also is DG to DF; [vi, 1] 

therefore DG is incommensurable with DF. [x. n] 

And both are rational ; 

therefore GD, DF are rational straight lines commensurable 
in square only. 

Therefore FG is an apotome. [x. 73] 

And FH is rational ; 

but the rectangle contained by a rational straight line and an 
apotome is irrational, [deduction from x. 10] 

and its "side" is irrational. 

And AC is the "side" of FE ; 

therefore AC is irrational. 

And let it be called that which produces with a 
medial area a medial whole. 

Q. E. D. 



x. 78, 79] PROPOSITIONS 78, 79 167 

In this case x, y have respectively the forms [cf. x. 35] 
pX* / ~k~~ p\* / k 

Suppose that x* i-y* = <ru, 

txy = ov. 

By hypothesis, the areas <ru, av are medial ; 
therefore w, v are both rational and ». <r (i). 

Further au v <rv, 
so that u ~j v , {*). 

Hence [(1), (2)] u, v are rational and **- , 
so that (« -») is the irrational straight line called apotome [x. 73]. 

Thus <r (« - u) is an irrational area, 
so that (x-yf, and consequently {.r-^), is irrational. 

The irrational straight line 

pA* 






is called that which products [i.e. when a square is described on it] with a 
medial area a media/ whole, more literally that which with a medial area makes 
the whole medial (15 /i<ri p-iaov fUvov to Skov jtoiouo-o). A clearer phrase (to 
us) would be the "side" of the difference between two medial areas, correspond- 
ing to the " side" of {the sum of) two mtdial areas [x, 4t], 



Proposition 79. 

To an apotome only one rational straight line can be 
annexed which is commensurable with the whole in square only. 

Let AB be an apotome, and BC an annex to it ; 
therefore AC, CB are rational 

straight lines commensurable in B c 

square only. [x. 73] — ' ' 

I say that no other rational 



straight line can be annexed to AB which is commensurable 
with the whole in square only. 

For, it" possible, let BD be so annexed ; 
therefore AD, DB are also rational straight lines commen- 
surable in square only. [*■ 73] 

Now, since the excess of the squares on AD, DB over 
twice the rectangle AD, DB is also the excess of the squares 
on AC, CB over twice the rectangle AC, CB, 
for both exceed by the same, the square on AB, [11. 7] 



i68 BOOK X [x. 79, 80 

therefore, alternately, the excess of the squares on AD, DB 
over the squares on AC, CB is the excess of twice the rect- 
angle AD, DB over twice the rectangle AC, CB. 

But the squares on AD, DB exceed the squares on AC, 
CB by a rational area, 
for both are rational ; 

therefore twice the rectangle AD, DB also exceeds twice the 
rectangle A C, CB by a rational area : 
which is impossible, 

for both are medial [x. 21], and a medial area does not exceed 
a medial by a rational area. [x. 26] 

Therefore no other rational straight line can be annexed 
to AB which is commensurable with the whole in square only. 

Therefore only one rational straight line can be annexed 
to an apotome which is commensurable with the whole in 
square only. 

Q. E. D. 

This proposition proves the equivalent of the well-known theorem of surds 
that, 

if a — Jb = x — ,Jy, then a - x, t=y; 
and, if Ja- Je = Jx~ Jy, then a = x, b =y. 

The method of proof corresponds to that of X. 42 for positive signs. 

Suppose, if possible, that an apotome can be expressed as (x -y) and also 
as (*'— y'), where x, y are rational straight lines commensurable in square only, 
and *', / are so also. 

Of x, x\ let x be the greater. 

Now, since x—y = x —y' % 

3? +y - (x" +/*) = ixy - zx'y. 

But {2? +y), (x* +y*) are both rational, so that their difference is a 
rational area. 

On the other hand, nxy, xx'y' are both medial areas, being of the form 

therefore the difference between two medial areas is rational : 
which is impossible [x. 26]. 
Therefore etc 

Proposition 80. 

To a first apotome of a medial straight line only one 
medial straight line can be annexed which is commensurable 
vnth the whole in square only and which contains with the 
whole a rational rectangle. 



x. 8o] PROPOSITIONS 79, 80 169 

For let AB be a first apotome of a medial straight line, 
and let BC be an annex to AB ; 

therefore AC, CB are medial ? £— - 

straight lines commensurable in 

square only and such that the rectangle AC, CB which they 
contain is rational ; [x. 74] 

I say that no other medial straight line can be annexed to 
AB which is commensurable with the whole in square only 
and which contains with the whole a rational area. 

For, if possible, let DB also be so annexed ; 
therefore AD, DB are medial straight lines commensurable 
in square only and such that the rectangle AD, DB which 
they contain is rational. [x. 74] 

Now, since the excess of the squares on AD, DB over 
twice the rectangle AD, DB is also the excess of the squares 
on AC, CB over twice the rectangle AC, CB, 

for they exceed by the same, the square on AB, [n. 7] 

therefore, alternately, the excess of the squares on AD, DB 
over the squares on AC, CB is also the excess of twice the 
rectangle AD, DB over twice the rectangle AC, CB. 

But twice the rectangle AD, DB exceeds twice the rect- 
angle AC, CB by a rational area, 
for both are rational. 

Therefore the squares on AD, DB also exceed the squares 
on A C, CB by a rational area : 

which is impossible, 

for both are medial [x. 15 and 23, Por.], and a medial area does 

not exceed a medial by a rational area. [x. 26] 

Therefore etc. 

Q. E. D. 

Suppose, if possible, that the same first apotome of a medial straight line 
can be expressed in terms of the required character in two ways, so that 

x-y = x'-y\ 
and suppose that x > x'. 

In this case x* +y', (x 11 +/*) are both media/ areas, and txy, ix'y 1 are both 
rational areas ; 
and x I +y t - (x" 1 +/') = ixy - ix'y'. 

Hence X. z6 is contradicted again ; 
therefore etc. 



170 



BOOK X 



[x. 81 






Proposition 81. 



C D 



To a second apotome of a medial straight line only one 
medial straight line can be annexed which is commensurable 
with the whole in square only and which contains with the 
whole a medial rectangle. 

Let AB be a second apotome of a medial straight line 
and BC an annex to AB ; 
therefore AC, CB are medial straight 
lines commensurable in square only and 
such that the rectangle AC, CB which 
they contain is medial. fi. 7s] 

I say that no other medial straight line 
can be annexed to AB which is commen- 
surable with the whole In square only and 
which contains with the whole a medial 
rectangle. 

For, if possible, let BD also be so 
annexed ; 

therefore AD, DB are also medial straight 
lines commensurable in square only and 
such that the rectangle AD, DB which 
they contain is medial. [x, 75] 

Let a rational straight line EF be set out, 

let EG equal to the squares on AC, CB be applied to EF, 
producing EM as breadth, 

and let HG equal to twice the rectangle AC, CB be sub- 
tracted, producing HM as breadth ; 

therefore the remainder EL is equal to the square on AB, 

[»• 7] 
so that AB is the "side" of EL. 

Again, let EF equal to the squares on AD, DB be applied 
to EF, producing EN as breadth. 

But EL is also equal to the square on AB ; 

therefore the remainder HI is equal to twice the rectangle 
AD, DB. [11. 7] 

Now, since AC, CB are medial straight lines, 
therefore the squares on AC, CB are also medial. 




x, 8i] PROPOSITION 81 lyi 

And they are equal to EG ; 

therefore EG is also medial. [x. 15 and 23, Por.] 

And it is applied to the rational straight line EF, producing 
EM as breadth ; 

therefore EM is rational and incommensurable in length 
with EF, [x. 23] 

Again, since the rectangle AC, CB is medial, 
twice the rectangle AC, CB is also medial. [x. *3, Por,] 

And it is equal to HG ; 

therefore HG is also medial. 

And it is applied to the rational straight line EF, producing 
HM as breadth ; 

therefore HM is also rational and incommensurable in length 
with EF. [x. j*] 

And, since AC, CB are commensurable in square only, 

therefore AC is incommensurable in length with CB. 

But, as A C is to CB, so is the square on AC to the rect- 
angle AC, CB; 

therefore the square on AC is incommensurable with the 
rectangle AC, CB. [x. 11] 

But the squares on AC, CB are commensurable with the 
square on AC, 

while twice the rectangle AC, CB is commensurable with the 
rectangle AC, CB ; [x. 6] 

therefore the squares on AC, CB are incommensurable with 
twice the rectangle AC, CB. [x. 13] 

And EG is equal to the squares on AC, CB, 
while GH is equal to twice the rectangle AC, CB ; 
therefore EG is incommensurable with HG. 

But, as EG Is to HG, so is EM to HM; [vi. 1] 

therefore EM is incommensurable in length with MH. [x. it] 

And both are rational ; 
therefore EM, MH are rational straight Hnes commensurable 
in square only ; 
therefore EH is an apotome, and HM an annex to it [x. 73J 



17* BOOK X [x. 81, 81 

Similarly we can prove that HN is also an annex to it ; 
therefore to an apotome different straight lines are annexed 
which are commensurable with the wholes in square only : 
which is impossible. [x. 79] 

Therefore etc. 

Q. E. O. 

As the irrationality of the setond apotome of a medial straight line was 
deduced [x. 75] from the irrationality of an apotome, so the present theorem 
is reduced to x. 79. 

Suppose, if possible, that (x -y), (x' -y') are the same second apotome of 
a medial straight line ; 
and let (say) x be greater than x'. 

Apply (Jf'+y), zxy and also (x* +y'*), 2x'y to a rational straight line a, 
i.e. put 

^•-"l and ^^r™',]. 
txy = <rv ) 2xy = av ) 

Dealing with (x -y) first, we have : 
{-v- +y") is a medial area, and ixy is also a medial area. 

Therefore u, v are both rational and ^ <r (1), 

Also, since x «-_y, x «_y, 
so that x 1 ^ xy, 
whence, as usual, x* +_y' w txy, 
that is, «u „ av, 
and therefore u v v (s). 

Thus [(1) and {2)] u, v are rational and «-, 
so that (it — v) is ah apotome. 

Similarly (u - if) is proved to be the same apotome. 

Hence this apotome is formed in two ways : 
which contradicts X. 79. 

Therefore the original hypothesis is false, and a sttond apotome of a 
media/ straight line is uniquely formed. 



Proposition 82. 

To a minor straight line only one straight line can be 
annexed which is incommensurable in square with the whole 
and which makes, with the whole, the sum 0/ the squares on 
them rational but twice the rectangle contained by them medial. 

Let AB be the minor straight line, and let BC be an 
annex to AB ; 

therefore AC, CB are straight * % ?_? 

lines incommensurable in square 



x. 8 2 , 83] PROPOSITIONS 81—83 *73 

which make the sum of the squares on them rational, but 
twice the rectangle contained by them medial. [x. 76] 

I say that no other straight line can be annexed to AB 
fulfilling the same conditions. 

For, if possible, let BD be so annexed ; 
therefore AD, DB are also straight lines incommensurable 
in square which fulfil the aforesaid conditions. [x. 76] 

Now, since the excess of the squares on AD, DB over 
the squares on AC, CB is also the excess of twice the rect- 
angle AD, DB over twice the rectangle AC, CB, 
while the squares on AD, DB exceed the squares on AC, 
CB by a rational area, 
for both are rational, 

therefore twice the rectangle AD, DB also exceeds twice 
the rectangle AC, CB by a rational area : 
which is impossible, for both are medial. [x. 36] 

Therefore to a minor straight line only one straight 
line can be annexed which is incommensurable in square with 
the whole and which makes the squares on them added 
together rational, but twice the rectangle contained by them 
medial. 

Q. E. D. 

Suppose, if possible, that, with the usual notation, 
x-y = x~y ; 
and let * (say) be greater than x. 

In this case (jc* +y>), (*-'* +y*) are both rational areas, 
and 2at, 2x y are both medial areas. 

But, as before, (x*+y*) — (*" +>'*) = xxy - zx'y', 
so that the difference between two medial areas is rational : 
which is impossible [x. 26]. 

Therefore etc. 



Proposition 83. 

To a straight line which produces with a rational area a 
medial whole only one straight line can be annexed which is 
incommensurable in square with the whole straight line and 
which with the whole straight line makes the sum 0/ the squares 
on them medial, but twice the rectangle contained by them 
rational. 



174 BOOK X fx. 83, 84 

Let AB be the straight line which produces with a rational 
area a medial whole, 

and let BC be an annex to AB ; A P S—° 

therefore AC, CB are straight lines 

incommensurable in square which fulfil the given conditions. 

I>- 77] 

I say that no other straight line can be annexed to AB 
which fulfils the same conditions. 

For, if possible, let BD be so annexed ; 
therefore AD, DB are also straight lines incommensurable in 
square which fulfil the given conditions. [x. 77] 

Since then, as in the preceding cases, 
the excess of the squares on AD, DB over the squares on 
AC, CB is also the excess of twice the rectangle AD, DB 
over twice the rectangle AC, CB, 

while twice the rectangle AD, DB exceeds twice the rectangle 
AC, CB by a rational area, 
for both are rational, 

therefore the squares on AD, DB also exceed the squares 
on AC, CB by a rational area : 
which is impossible, for both are medial. [x. 26] 

Therefore no other straight line can be annexed to AB 
which is incommensurable in square with the whole and which 
with the whole fulfils the aforesaid conditions; 
therefore only one straight line can be so annexed. 

Q. E. D. 

Suppose, with the same notation, that 

x —y = * ' — y . (x > x' ) 

Here, (x'+y 1 ), (x^+y'') being both medial areas, and 2xy, ix'y' both 
rational areas, 

while (x* +y*) — (i* 1 +y*) ■ zxy - ix'y, 

x. 26 is contradicted again. 

Therefore etc. 



Proposition 84. 

To a straight line which produces with a medial area a 
medial whole only one straight line can be annexed which is 
incommensurable in square with the whole straight line and 
which with the whole straight line makes the sum of the squares 



*.&»] 



PROPOSITIONS 83, 84 



I7S 



on them medial and twice the rectangle contained by them both 
medial and also incommensurable with the sum of the squares 
on them. 






Let AB be the straight line which produces with a medial 
area a medial, whole, 
and BC an annex to it ; 

therefore AC, t"Z?are straight lines incommensurable in square 
which fulfil the aforesaid conditions. [x. 7 8 ] 



c D 



E H 










F L 



I say that no other straight line can be annexed to AB 
which fulfils the aforesaid conditions. 

For, if possible, let BD be so annexed, 

so that AD, DB are also straight lines incommensurable in 
square which make the squares on AD, DB added together 
medial, twice the rectangle AD, DB medial, and also the 
squares on AD, DB incommensurable with twice the rectangle 
AD, DB, [x. 7 «] 

Let a rational straight line EF be set out, 

let EG equal to the squares on AC, CB be applied to EF, 
producing EM as breadth, 

and let HG equal to twice the rectangle AC, CB be applied 
to EF, producing HM as breadth ; 

therefore the remainder, the square on AB [11. 7], is equal 

to EL ; 

therefore AB is the "side" of EL. 






Again, let EI equal to the squares on AD, DB be applied 
to EF, producing EN as breadth. 

But the square on AB is also equal to EL ; 

therefore the remainder, twice the rectangle AD, DB [n. 7], 
is equal to HI. 



176 BOOK X [x. 84 

Now, since the sum of the squares on AC, CB is medial 
and is equal to EG, 
therefore EG is also medial. 

And it is applied to the rational straight line EF, pro- 
ducing EM as breadth ; 

therefore EM is rational and incommensurable in length 
with EF. [x. it] 

Again, since twice the rectangle AC, CB is medial and is 
equal to HG, 

therefore HG is also medial. 

And it is applied to the rational straight line EF, pro- 
ducing HM as breadth ; 

therefore HM is rational and incommensurable in length 
with EF. [x. as] 

And, since the squares on AC, CB are incommensurable 
with twice the rectangle AC, CB, 

EG is also incommensurable with HG ; 

therefore EM is also incommensurable in length with MH. 

[vi. i, x. 1 1] 
And both are rational ; 

therefore EM, MH are rational straight lines commensurable 
in square only ; 

therefore EH is an apotome, and HM an annex to it. [x. 73] 

Similarly we can prove that EH is again an apotome and 
HN an annex to it. 

Therefore to an apotome different rational straight lines 
are annexed which are commensurable with the wholes in 
square only : 

which was proved impossible. [x, 79J 

Therefore no other straight line can be so annexed to AB. 

Therefore to AB only one straight line can be annexed 
which is incommensurable in square with the whole and which 
with the whole makes the squares on them added together 
medial, twice the rectangle contained by them medial, and 
also the squares on them incommensurable with twice the 
rectangle contained by them. 

Q. E, D. 



X.84] PROPOSITION 84, DEFINITIONS III. 177 

With the usual notation, suppose that 

x-y = x"-?. (■*>*') 

Let x'+y^em ] , x" , +y* = <ru' 1 

} and "J , , } . 

Consider (x-y) first; 
it follows, since (x* +y), 2xy are both medial areas, that 
u, v are both rational and ■j <r (1). 

But x* +y >j axy, 
that is, au ^ <ni, 
and therefore M « v (a). 

Therefore [(1) and (2)] v, v are rational and «- ; 
hence (u-v) is an apotome. 

Similarly (*' - r/) is proved to be the j»»w apotome. 

Thus the same apotome is formed as such in two ways : 
which is impossible [x. 79], 

Therefore, etc. 

DEFINITIONS III. 

1 . Given: a rational straight line and an apotome, if the 
square on the whole be greater than the square on the annex 
by the square on a straight line commensurable in length with 
the whole, and the whole be commensurable in length with 
the rational straight line set out, let the apotome be called a 
first apotome. 

2. But if the annex be commensurable in length with 
the rational straight line set out, and the square on the whole 
be greater than that on the annex by the square on a straight 
line commensurable with the whole, let the apotome be called 
a second apotome. 

3. But if neither be commensurable in length with the 
rational straight line set out, and the square on the whole be 
greater than the square on the annex by the square on a 
straight line commensurable with the whole, let the apotome 
be called a third apotome, 

4. Again, if the square on the whole be greater than 
the square on the annex by the square on a straight line 
incommensurable with the whole, then, if the whole be com- 
mensurable in length with the rational straight line set out, 
let the apotome be called a fourth apotome ; 

5. if the annex be so commensurable, a fifth ; 

6. and, if neither, a sixth. 



r 7 8 BOOK X [x. 85 

Proposition 85. 

To find the first apotome. 

Let a rational straight line A be set out, 
and let BG be commensurable in length with A ; 
therefore BG is also rational. 






A- 
H- 



E F D 

Let two square numbers DE, EF be set out, and let their 
difference FD not be square ; 

therefore neither has ED to DF the ratio which a square 
number has to a square number. 

Let it be contrived that, 

as ED is to DF, so is the square on BG to the square on GC; 

[x. 6, Por.] 
therefore the square on BG is commensurable with the square 
on GC. [x. 6] 

But the square on BG is rational ; 
therefore the square on GC is also rational ; 
therefore GC is also rational. 

And, since ED has not to DF the ratio which a square 
number has to a square number, 

therefore neither has the square on BG to the square on GC 
the ratio which a square number has to a square number ; 
therefore BG is incommensurable in length with GC. [x. 9] 

And both are rational ; 
therefore BG, GC are rational straight lines commensurable 
in square only ; 
therefore BC is an apotome, [x. 73] 

I say next that it is also a first apotome. 

For let the square on H be that by which the square on 
BG is greater than the square on GC. 

Now since, as ED is to FD, so is the square on BG to 
the square on GC, 

therefore also, convertendo, [v. 19, Por.] 

as DE is to EF, so is the square on GB to the square on H. 



x. 8 S ] PROPOSITION 8 S 179 

But DE has to EF the ratio which a square number has 
to a square number, 

for each is square ; 

therefore the square on GB also has to the square on H the 
ratio which a square number has to a square number ; 

therefore BG is commensurable in length with H. [x. 9] 

And the square on BG is greater than the square on GC 
by the square on H \ 

therefore the square on BG is greater than the square on GC 
by the square on a straight Tine commensurable in length 
with BG. 

And the whole BG is commensurable in length with the 
rational straight line A set out. 

Therefore BC is a first apotome. [x. Deff. in. 1] 

Therefore the first apotome BC has been found. 

(Being) that which it was required to find. 

Take kp commensurable in length with p, the given rational straight line. 

Let m\ «' be square numbers such that (« , -« 1 ) is not square. 

Take * such that m* : (n* -*) = #? : x 1 (1), 

so that x = kp — — ■ ■ 

= kpji-H, say. 
Then shall ip-x, or kp -kp -/i-X 1 , be a. first apotome. 
For (a) it follows rrom (i) that x is rational but incommensurable with kp, 
whence kp, x are rational and <*-» 
so that (kp—x) is an apotome. 
(ft) lt/ = *Y - x*, then, by ( 1), toMwrtenJo, 

m'-.n'^Pp'.y, 
whence y, that is, <Jk*p* - .**, is commensurable in length with kp. 

And kp " p ; 
therefore kp-xh & first apotome. 

As explained in the note to x. 48, the first apotome 
kp — kp>Ji -A' 
is one of the roots of the equation 

x t -iAp.x + k 1 & t ( r = o. 






i8o BOOK X [x.86 

Proposition 86. 
To find the second apotome. 

Let a rational straight line A be set out, and GC com- 
mensurable in length with A ; 
therefore GC is rational. 

Let two square numbers DE, ? — £_ — ■ — ? 

EF be set out, and let their h 

difference DF not be square. 

Now let it be contrived that, g f b 

as FD is to DE, so is the square 

on CG to the square on GB. [x. 6, Por.] 

Therefore the square on CG is commensurable with the 
square on GB. [x. 6] 

But the square on CG is rational ; 
therefore the square on GB is also rational ; 
therefore BG is rational. 

And, since the square on GC has not to the square on GB 
the ratio which a square number has to a square number, 
CG is incommensurable in length with GB, [x, 9] 

And both are rational j 
therefore CG, GB are rational straight lines commensurable 
in square only ; 
therefore EC is an apotome. [x. 73] 

I say next that it is also a second apotome. 

For let the square on H be that by which the square on 
BG is greater than the square on GC. 

Since then, as the square on BG is to the square on GC, 
so is the number ED to the number DF, 
therefore, convertendo, 

as the square on BG is to the square on H, so is DE to EF. 

[v. 19, Por.] 

And each of the numbers DE, EF is square ; 
therefore the square on BG has to the square on // the ratio 
which a square number has to a square number ; 
therefore BG is commensurable in length with H. [x. g] 

And the square on BG is greater than the square on GC 
by the square on H ; 
therefore the square on BG is greater than the square on GC 



x. 86, 87] PROPOSITIONS 86, 87 i8r 

by the square on a straight line commensurable in length 
with BG. 

And CG, the annex, is commensurable with the rational 
straight line A set out. 

Therefore BC is a second apotome. [x. Deff. 111. 2] 

Therefore the second apotome BC has been found. 

Q. E. D. 
Take, as before, kp commensurable in length with p. 
Let m\ n 1 be again square numbers, but («' - « s ) not square. 
Take * such that («*-«•) ;*** = j&y :.*• (1), 

whence x = kp , 

V *>' — »' 
kp 

Thus x is zrcater than kp. 
kp 

Then x-ip, or . — kp, is a second apotome, 

Vi-X* 

For (a), as before, x is rational and «- V- 
fj9) If «" - #p' ■ y, we have, from ( 1 ), 

m* :n*~x' :^. 
Thus,?, or Jxt-Pp*, is commensurable in length with x. 
And kp is f> p. 

Therefore x — kp is a second apotome. 
As explained in the note on x. 49, the second apotome 

is the lesser root of the equation 

Proposition 87. 

To find the third apotome. 

Let a rational straight line A be set out, 
let three numbers E, BC, CD be 

set out which have not to one 

another the ratio which a square ? - tj . 

number has to a square number, 

but let CB have to BD the ratio 

which a square number has to a e 

square number. . 

Let it be contrived that, as E B D 

is to BC, so is the square on A to the square on FG, 






i8j BOOK X [x. 87 

and, as BC is to CD, so is the square on FG to the square 
on GH. [* 6, Por.] 

Since then, as E is to BC, so is the square on A to the 
square on FG, 

therefore the square on A is commensurable with the square 
on FG. [* 6] 

But the square on A is rational ; 
therefore the square on FG is also rational ; 
therefore FG is rational. 

And, since E has not to BC the ratio which a square 
number has to a square number, 

therefore neither has the square on A to the square on FG 
the ratio which a square number has to a square number ; 
therefore A is incommensurable in length with FG, [x. 9] 

Again, since, as BC is to CD, so is the square on FG to 
the square on GH, 

therefore the square on FG is commensurable with the square 
on GH. [x. 6] 

But the square on FG is rational ; 
therefore the square on GH is also rational ; 
therefore GH is rational. 

And, since BC has not to CD the ratio which a square 
number has to a square number, 

therefore neither has the square on FG to the square on GH 
the ratio which a square number has to a square number ; 
therefore FG is incommensurable in length with GH. [x. 9] 

And both are rational ; 
therefore FG, GH are rational straight lines commensurable 
in square only ; 

therefore FH is an apotome. [x. 73] 

I say next that it is also a third apotome. 
For since, as E is to BC, so is the square on A to the 
square on FG, 

and, as BC is to CD, so is the square on FG to the square 
on HG, 

therefore, ex aequali, as E is to CD, so is the square on A 
to the square on HG. [v, %%\ 



x, 8 7 ] PROPOSITION 87 183 

But E has not to CD the ratio which a square number 
has to a square number ; 

therefore neither has the square on A to the square on GH 
the ratio which a square number has to a square number ; 

therefore A is incommensurable in length with GH. [x. 9] 

Therefore neither of the straight lines FG, GH is 
commensurable in length with the rational straight line A 
set out. 

Now let the square on AT be that by which the square on 
FG is greater than the square on GH. 

Since then, as BC is to CD, so is the square on FG to 
the square on GH, 

therefore, convertendo, as BC is to BD, so is the square on 
FG to the square on K. [v. 19, Por.] 

But BC has to BD the ratio which a square number has 
to a square number ; 

therefore the square on FG also has to the square on K the 
ratio which a square number has to a square number. 

Therefore FG is commensurable in length with K, [x. 9] 

and the square on FG is greater than the square on GH by 
the square on a straight line commensurable with FG. 

And neither of the straight lines FG, GH is commen- 
surable in length with the rational straight line A set out ; 
therefore FH is a third apotome. [*• Deff. m. 3] 

Therefore the third apotome FH has been found. 

Q. E. D. 

Let p be a rational straight line. 

Take numbers p, qm", a (m* - *») which hare not to one another the ratio 
of square to square. 

Now let x, y be such that 

P:qm' = ?:x* (i) 

and ?«' :q (^w , -» , ) = ^ , -.y (a). 

Then shall (* —y\ be a tkird apotome. 
For (a), from (i), 

* is rational but v p (3), 

And, from (z),y is rational but « x. 
Therefore x, y are rational and ^-, 
so that {x -y) is an apotome. 



.84 BOOK X [x. 87, 88 

(0) By (1), (j), ex aequali, 

p\q(m*-#) = p*:f, 
whence y w p. 

Thus, by this and (3), x, y are both vp (4)- 

Lastly, let ** = ** —y*, so that, from (1), eomxrteitdo, 
gm* : qn' = x* : **; 

therefore *, or *Jx*-y*, n x {5). 

Thus [(4) and (5)] (x-y) is a third apotome. 
To find its form, we have, from (1) and (2), 

J"p- — -jp> 

so that *"■' = ^/l'^*" , '" |l ""'' , 

This may be written in the form 

mjk.p- mjk . pJi -X*. 
As explained in the note on x. 50, this is the lesser root of the equation 
x*— 2MjJk.px + k t m i ibp t = 0. 



Proposition 88. 
To find the fourth apotome. 

Let a rational straight line A be set out, and BG com- 
mensurable in length with it ; 
therefore BG is also rational. 

A ? $ ? 



H- 



Let two numbers DF, FE be set out such that the whole 
DE has not to either of the numbers DF, EF the ratio 
which a square number has to a square number. 

Let it be contrived that, as DE is to EF, so is the square 
on BG to the square on GC ; [x. 6, Pot.] 

therefore the square on BG is commensurable with the square 
on GC. [x, 6] 

But the square on BG is rational ; 
therefore the square on GC is also rational ; 
therefore GC is rational. 



x. 88] PROPOSITIONS 87, 88 185 

Now, since DE has not to EF the ratio which 3 square 
number has to a square number, 

therefore neither has the square on BG to the square on GC 
the ratio which a square number has to a square number ; 
therefore BG is incommensurable in length with GC. [x. 9] 

And both are rational ; 
therefore BG, GC are rational straight lines commensurable 
in square only ; 
therefore BC is an apotome. [x. 73] 

Now let the square on H be that by which the square on 
BG is greater than the square on GC. 

Since then, as DE is to EF, so is the square on BG to 
the square on GC, 

therefore also, cmvertmdo, as ED is to DF, so is the square 
on GB to the square on H. [v. 19, Por.] 

But ED has not to DF the ratio which a square number 
has to a square number ; 

therefore neither has the square on GB to the square on H 
the ratio which a square number has to a square number ; 

therefore BG is incommensurable in length with H. [x. 9] 

And the square on BG is greater than the square on GC 
by the square on H; 

therefore the square on BG is greater than the square on GC 
by the square on a straight line incommensurable with BG. 

And the whole BG is commensurable in length with the 
rational straight line A set out 

Therefore BC is a fourth apotome. [x. Deff, in. 4] 

Therefore the fourth apotome has been found. 

Q. £. D. 
Beginning with p, k P , as in x. 85, 86, we take numbers m, n such that 
(m + n) has not to either of the numbers m, n the ratio of a square number to 
a square number. 

Take x such that (m + *) : n = ¥p' : x* ...(1), 

whence * = ip* / 

Then shall (tp-x), or (ip r 4— T ) , be & fourth aPctome. 

\ Vi + A/ 



t86 



BOOK X 



[x. 88, 89 






For, by (1), x is rational and *» kpt. 
Also *J&f?~x* is incommensurable with kp, since 
(» + ») :« = £y :{*y-*»), 

and the ratio (« + *): m is not that of a square number to a square number. 
And kp p. 
As explained in the note on x, 51, ihtfottrtA apotome 

is the lesser root of the quadratic equation 

X* - 2kp , X + - *V = O. 

r 1 + A r 






Proposition 89. 
To find the fifth apotome. 

Let a rational straight line A be set out, 
and let CG be commensurable in length 
with A ; 

therefore CG is rational. 

Let two numbers DF, FE be set out 
such that DE again has not to either of the 
numbers DF, FE the ratio which a square 
number has to a square number ; 
and let it be contrived that, as FE is to ED, 
so is the square on CG to the square on GB. 

Therefore the square on GB is also 
rational ; [x. 6] 

therefore BG is also rational. 

Now since, as DE is to EF, so is the square on BG to 
the square on GC, 

while DE has not to EF the ratio which a square number 
has to a square number, 

therefore neither has the square on BG to the square on GC 
the ratio which a square number has to a square number ; 
therefore BG is incommensurable in length with GC. fx. 9] 

And both are rational ; 
therefore BG, GC are rational straight lines commensurable 
in square only ; 
therefore BC is an apotome. [x. 73] 



x. 89] PROPOSITIONS 88, 89 187 

I say next that it is also a fifth apotome. 

For let the square on H be that by which the square on 
BG is greater than the square on GC. 

Since then, as the square on BG is to the square on GC, 
so is DE to EF, 

therefore, eonverlendo, as ED is to DF, so is the square on 
BG to the square on H. [v. 19, Por.] 

But ED has not to DF the ratio which a square number 
has to a square number ; 

therefore neither has the square on BG to the square on H 
the ratio which a square number has to a square number ; 

therefore BG is incommensurable in length with H. [x. 9] 

And the square on BG is greater than the square on GC 
by the square on H \ 

therefore the square on GB is greater than the square on GC 
by the square on a straight line incommensurable in length 
with GB. 

And the annex CG is commensurable in length with the 
rational straight line A set out ; 

therefore BC is a fifth apotome. [x. Deff. m. 5] 

Therefore the fifth apotome BC has been found. 

Q. E, D. 

Let p, ip and the numbers m, n of the last proposition be taken. 

Take x such that »:(*» + *) = *V : x* (1). 

In this case x> kp, and x = kp,/ 

= kp*fi +A, say. 

Then shall (x - kp), or (kpji + A — kp), be a, fifth apotome. 

For, by (1), x is rational and «- kp. 

And since, by (1), (m + ») : m m x* : (x* - k*p*), 

1/3^- k*f? is incommensurable with x. 

Also kp « p. 

As explained in the note on x. 51, Ihefi/tk apotome 

kp*ft+k- kp 

is the lesser root of the quadratic 

** - ikpjl + X . X + XyP/>' = o. 



BOOK X [x. 90 



Proposition 90. 
To find the sixth apotome. 

Let a rational straight line A be set out, and three 
numbers E, BC, CD not having 

to one another the ratio which A 

a square number has to a square f * - a 

number ; 

and further let CB also not have 

to BD the ratio which a square 

number has to a square number. g g c 

Let it be contrived that, as 
E is to BC, so is the square on A to the square on FG, 

and, as BC is to CD, so is the square on FG to the square 
on GH. [x. 6, For.] 

Now since, as E is to BC, so is the square on A to the 
square on FG, 

therefore the square on A is commensurable with the square 
on FG. [x. 6] 

But the square on A is rational ; 

therefore the square on FG is also rational ; 

therefore FG is also rational. 

And, since E has not to BC the ratio which a square 
number has to a square number, 

therefore neither has the square on A to the square on FG 
the ratio which a square number has to a square number ; 
therefore A is incommensurable in length with FG. [x. 9] 

Again, since, as BC is to CD, so is the square on FG to 
the square on GH, 

therefore the square on FG is commensurable with the square 
on GH. [x. 6] 

But the square on FG is rational ; 

therefore the square on GH is also rational ; 

therefore GH is also rational. 

And, since BC has not to CD the ratio which a square 
number has to a square number, 



x. 90] PROPOSITION 90 189 

therefore neither has the square on FG to the square on GH 
the ratio which a square number has to a square number ; 
therefore FG is incommensurable in length with GH. [x. 9] 

And both are rational ; 
therefore FG, GH are rational straight lines commensurable 
in square only ; 
therefore FH is an apotome. [x. 73] 

I say next that it is also a sixth apotome. 

For since, as E is to BC, so is the square on A to the 
square on FG, 

and, as BC is to CD, so is the square on FG to the square 
on GH, 

therefore, ex aeguaii, as E is to CD, so is the square on A to 
the square on GH. [v, *i] 

But E has not to CD the ratio which a square number 
has to a square number ; 

therefore neither has the square on A to the square on GH 
the ratio which a square number has to a square number ; 

therefore A is incommensurable in length with GH; [x. 9] 

therefore neither of the straight lines FG, GH is commen- 
surable in length with the rational straight line A. 

Now let the square on A!" be that by which the square on 
FG is greater than the square on GH. 

Since then, as BC is to CD, so is the square on FG to 
the square on GH, 

therefore, conver tench, as CB is to BD, so is the square on 
FG to the square on K. [v. 19, Por.] 

But CB has not to BD the ratio which a square number 
has to a square number ; 

therefore neither has the square on FG to the square on K 
the ratio which a square number has to a square number ; 

therefore FG is incommensurable in length with K. [x. 9] 

And the square on FG is greater than the square on GH 
by the square on K ; 

therefore the square on FG is greater than the square on GH 
by the square on a straight line incommensurable in length 
with FG. 



iqo BOOK X [x. 90, 91 

And neither of the straight lines FG, GH is commen- 
surable with the rational straight line A set out. 

Therefore FH is a sixth apotome. [x. Deff. 111. 6] 

Therefore the sixth apotome FH has been found. 

Q. E. D. 

Let p be the given rational straight line. 

Take numbers p, (m + it), n which have not to one another the ratio of a 
square number to a square number, m, n being also chosen such that the 
ratio ( m + n) ; m is not that of square to square. 

Take x,y such that / : (» + «) = p*: x* (i), 

(»! + n) : n = x t :f (?). 

Then shall (x -y) be a sixth apotome. 

For, by ( i ), x is rational and w p (3). 

By {3), since x is rational, 

y is rational and « x (4). 

Thus [(3), (4)] (x -y) is an apotome. 
Again, ex atquali, p : n = p* : _v*, 

whence y - p. 

Thus x, y are both u p. 
Lastly, (onvertendo from (a), 

(m + «ri : m = «? : («* -^"), 
whence Jx* -y* ~ at. 

Therefore (x-y) is a rucrA apotome. 
From (1) and (2) we have 

fm + n 

so that the sixth apotomt may be written 

/m + n In 

p v ~r ~ p v ? ' 

or, more simply, Jk . p — ,/X . p. 

As explained in the note on x. 53, the j/#M apotome is the lesser root of 

the equation 

X s - i^* ■ P* + (* "* ty P* = * 

Proposition 91. 

//" «» area £« contained by a rational straight line and a 
first apotome, the "side" of the area is an apotome. 

For let the area AB be contained by the rational straight 
line AC and the first apotome AD ; 

I say that the "side" of the area AB is an apotome. 



X. gij 



PROPOSITIONS 90, 91 



191 



For, since AD is a first apotome, let DG be its annex ; 

therefore AG, GD are rational straight lines commensurable 
in square only. [x. 73] 

And the whole AG is commensurable with the rational 
straight line AC set out, 

and the square on AG is greater than the square on GD 
by the square on a straight line commensurable in length 
with AG; [x. Deff. m. 1] 

if therefore there be applied to AG a parallelogram equal to 
the fourth part of the square on DG and deficient by a square 
figure, it divides it into commensurable parts. [x. 17] 






A 


D 


E 




p a 










C 

L 

6 


B N 




K 








U/ 

/ W 






1 


i 




■ 


^ 


1 






Let DG be bisected at E, 

let there be applied to AG a parallelogram equal to the square 
on EG and deficient by a square figure, 
and let it be the rectangle AF, FG ; 
therefore AF is commensurable with FG. 

And through the points E, F, G let EH, FI^ GK be drawn 
parallel to AC, 

Now, since AF is commensurable in length with FG, 

therefore AG is also commensurable in length with each of 
the straight lines AF, FG. [x. 15] 

But AG is commensurable with AC; 

therefore each of the straight lines AF, FG is commensurable 
in length with AC. [x. ia] 



i 9 * BOOK X [x. 91 

And AC is rational ; 
therefore each of the straight lines AF, FG is also rational, 
so that each of the rectangles A I, FK is also rational, [x. 19] 

Now, since DE is commensurable in length with EG, 

therefore DG is also commensurable in length with each of 
the straight lines DE, EG. [x. 15] 

But DG is rational and incommensurable in length 
with AC; 

therefore each of the straight lines DE, EG is also rational 
and incommensurable in length with AC] [x. *3] 

therefore each of the rectangles DH, EK is medial. [x. u] 

Now let the square LM be made equal to A I, and let 
there be subtracted the square NO having a common angle 
with it, the angle LPM, and equal to FK; 

therefore the squares LM, NO are about the same diameter. 

[vi. 26] 

Let PR be their diameter, and let the figure be drawn. 
Since then the rectangle contained by AF, FG is equal to 

the square on EG, 

therefore, as AF is to EG, so is EG to FG. [vi. 17] 

But, as AF is to EG, so is A I to EK, 

and, as EG is to FG, so is EK to KF; [vi. 1] 

therefore EKis a mean proportional between A J, KF. [v. 1 1] 

But MN is also a mean proportional between LM, NO, 
as was before proved, [Lemma after x, 53] 

and AI is equal to the square LM, and KF to NO ; 

therefore MN is also equal to EK. 

But EK is equal to DH, and MN to LO ; 
therefore DK is equal to the gnomon UVW a.n& NO. 

But AK is also equal to the squares LM, NO ; 
therefore the remainder AB is equal to ST. 

But ST is the square on LN ; 
therefore the square on LN is equal to AB ; 
therefore LN is the "side" of AB, 



x, 91] PROPOSITION 91 193 

I say next that LN is an apotome. 

For, since each of the rectangles A I, FK is rational, 
and they are equal to LM, NO, 

therefore each of the squares LM, NO, that is, the squares on 
LP, PN respectively, is also rational ; 
therefore each of the straight lines LP, PN is also rational. 

Again, since DH is medial and is equal to LO, 
therefore LO is also medial. 

Since then LO is medial, 
while NO is rational, 
therefore LO is incommensurable with NO. 

But, as LO is to NO, so is LP to PN; [vi. i] 

therefore LP is incommensurable in length with PN. [x. 1 1] 

And both are rational ; 
therefore LP, PN are rational straight lines commensurable 
in square only ; 
therefore LN is an apotome. [x. 73] 

And it is the "side" of the area AB ; 
therefore the "side" of the area AB is an apotome. 

Therefore etc. 

This proposition corresponds to x. 54, and the problem solved in it is to 
find and to classify the side of a square equal to the rectangle contained by a 
first apotome and p, or (algebraically) to find 

Jp{kp-kpJ7=%). 
First find u, v from the equations 

' . ' U ■¥ V = kp 

»» = 1W(i-V), 
If u, v represent the values so found, put 

y = pvi w 

and (x —y) shall be the square root required. 

To prove this Euclid argues thus. 

By (1), u : \kp Vi^V = \kp <J t~k* ; v, 

whence ptt ! J ip* Ji -X s = J kp 1 Ji -A* : pv, 

or x* : i V VF^A" = i V Vr^X" : f. 

But [Lemma after x. 53] 

X? ; xy = xy : y 1 , 
so that *ji = ivVi-A» (3). 



x-)} <* 



194 BOOK X [x. 91, 9* 

Therefore (x -y) 1 - x? + j> ! - 2xy 

-V-VVr-X*. 

Thus ^— ^) is equal to J p(kp - kp Ji —X'). 

It has next to be proved that (#-,y) is an apotome. 

From (1) it follows, by x. 17, that 

u " v; 
thus v, & are both commensurable with (u + v) and therefore with p (4), 

Hence u, v are both rational, 
so that pu, pv are rational areas ; 

therefore, by { 2), x\y* are rational and commensurable (5), 

whence also x, y are rational straight lines (6). 

Next, kp Vi - A* is rational and ^ p ; 
therefore ikp" 1 Ji -\' is a medial area. 

That is, by (3}, xy is a medial area. 

But [(5)] y is a rational area ; 
therefore jtv ^ y 1 , 

Or x \s y. 

But [(6)] *, jy are both rational. 

Therefore x, y are rational and f~ ; 
so that (x — y) is an apotome. 

To find the form of (x-y) algebraically, we have, by solving (1), 
u = hkp (1 + A), 
y = &lp(i-A), 

whence, from <*), * = p ^/- (1 + K), 

and *-J"»py -{» + A >-py J (»-*■> 

As explained in the note on x. 54, (*-^) is the lesser positive root of the 
biquadratic equation 

X t -*&p*.X* + VPp'=t>. 



Proposition 92. 

If an area be contained by a rational straight line and a 
second apotome, the "side" of the area is a first apotome of a 
medial straight line. 

For let the area AB be contained by the rational straight 
line AC and the second apotome AD ; 



X, 92 ] 



PROPOSITIONS 9i, 92 



*9S 



I say that the "side" of the area AB is a first apotome of a 
medial straight line. 











G~ " 


Q H 
N 


1 K 


L 
S 


W/~ 


f 


P 


/\i 


sp 




1 


I 






f. 


1 



For let DG be the annex to AD ; 

therefore AG, GD are rational straight lines commensurable 
in square only, [x. 73] 

and the annex DG is commensurable with the rational straight 
line AC set out, 

while the square on the whole AG is greater than the square 
on the annex GD by the square on a straight line commen- 
surable in length with AG. [x. Deffi 111. a] 

Since then the square on AG is greater than the square 
on GD by the square on a straight line commensurable 
with AG, 

therefore, if there be applied to AG a parallelogram equal to 
the fourth part of the square on GD and deficient by a square 
figure, it divides it into commensurable parts. [x. 17] 

Let then DG be bisected at F, 
let there be applied to AG a parallelogram equal to the square 
on EG and deficient by a square figure, 

and let it be the rectangle AF, FG ; 

therefore AF 'is commensurable in length with FG. 

Therefore AG is also commensurable in length with each 
of the straight lines AF, FG. [x. 15] 

But AG is rational and incommensurable in length 
with AC; 



r 9 6 BOOK X [x. 92 

therefore each of the straight lines AF, FG is also rational 
and incommensurable in length with AC; [x. 13] 

therefore each of the rectangles A J, FK is medial. [x. 1 1] 

Again, since DE is commensurable with EG, 
therefore DG is also commensurable with each of the straight 
lines DE, EG. [s. 15] 

But DG is commensurable In length with AC. 
Therefore each of the rectangles DH, EK is rational. 

[x. 19] 

Let then the square LM he constructed equal to A I, 
and let there be subtracted NO equal to FK and being about 
the same angle with LM, namely the angle LPM; 

therefore the squares LM, NO are about the same diameter. 

[vi. 26] 

Let PR be their diameter, and let the figure be drawn. 

Since then AI, FK&K medial and are equal to the squares 
on LP, PN, 
the squares on LP, PN are also medial ; 

therefore LP, PN are also medial straight lines commen- 
surable in square only. 

And, since the rectangle AF, EG is equal to the square 
on EG, 

therefore, as AF is to EG, so is EG to EG, [vi. 17] 

while, as AF is to EG, so is AI to EK, 

and, as EG is to EG, so is EK to FK ; [vi. 1] 

therefore EK is a mean proportional between AI,FK. [v. u] 

But MN is also a mean proportional between the squares 
LM, NO, 

and AI is equal to LM, and FK to NO ; 

therefore MN is also equal to EK. 

But DH is equal to EK, and L equal to MN ; 

therefore the whole DK is equal to the gnomon UVW 
and NO. 

Since then the whole AK is equal to LM, NO, 

and, in these, DK is equal to the gnomon UVW a,n& NO, 

therefore the remainder AB is equal to TS. 



x. ga] PROPOSITION 92 197 

But TS is the square on LN; 
therefore the square on LN is equal to the area AB ; 
therefore LN is the "side" of the area AB. 

I say that LN is a first apotome of a medial straight line. 
For, since EK is rational and is equal to LO, 

therefore LO, that is, the rectangle LP, PN, is rational. 

But NO was proved medial ; 
therefore L,0 is incommensurable with NO. 

But, as LO is to NO, so is LP to PN ; [vi. 1] 

therefore LP, PN are incommensurable in length. [x. 11] 

Therefore LP, PN are medial straight lines commen- 
surable in square only which contain a rational rectangle ; 

therefore LN is a first apotome of a medial straight line. 

[*■ 74] 
And it is the "side" of the area AB. 
Therefore the "side" of the area AB is a first apotome 
of a medial straight line. 

q. e. n. 

There is an evident flaw in the text in the place {Heiberg, p. 28*, 
11. 1 7 — ta : translation p. 1 96 above) where it is said that "since then A I, FK 
are medial and are equal to the squares on LP, PN, the squares on LP, PN 
are also medial ; therefore LP, PN are also medial straight lines commensurable 
in square only" It is not till the last lines of the proposition (Heiberg, p. 284, 
11. 17, 18) that it is proved that LP, /Ware incommensurabk in length. What 
should have been proved in the former passage is that the squares on LP, PN 
are commensurable, so that LP, PN are commensurable in square (not 
commensurable in square only). I have supplied the step in the note below : 
"Also x'^y', since a « v." Theon seems to have observed the omission and 
to have put "and commensurable with one another" after " medial" in the 
passage quoted, though even this does not show why the squares on LP, PN 
are commensurable. One ms. (V) also has "only" ()Uvoy) erased after 
"commensurable in square." 



This proposition amounts to finding and classifying 



The method is that of the last proposition. Euclid solves, first, the 
equations 

kp 

uv = |iy J 



i 9 S BOOK X [x. os 

Then, using the values of u, v so found, he puts 

& -pu 
f = pv 



■**■} <*>, 



and {x -y) is the square root required. 

is proved in the same way as is the corresponding fact in x. 91. 

From (1) u:\kp = \kp : v, 

so that pu : J &p* - \&f? : pv. 

But X? : xy = xy :y', 
whence, by (j), xy = \kp i {3)- 

Therefore- (* — ?)* -^ *^~ '^ 

= p(u + v)-kp* 



■'&-*)■ 






Next, we have to prove that {x — y) is a first apotomt of a medial straight 
line. 

From (1) it follows, by x, 17, that 

K" v , (4). 

therefore u, v are both <"> (« + v). 

But [(1)] (w + v) is rational and v p ; 
therefore tf, » are both rational and v p (5). 

Therefore pu, pv, or *", j* 1 , are both medial areas, and .v, j* are medial 
straight lines (6). 

Also 3? ~y, sinr_e » « w [(4)] (7). 

Now jy, or J ip', is a rational area ; 
therefore ^«/, 

and * ^ Jd 

Hence [(6), (7), (3)] #, ^ are medial straight lines commensurable in square 
only and containing a rational rectangle ; 

therefore {x-y) is & first apotomt of a medial straight line. 

Algebraical solution of the equations gives 

. 1 -X . 
» = 4 - - , *?■ 






and x 



->-v/[(^)'-,v/^)' 



As explained in the note on x. 55, this is the lesser positive root of the 
equation 



X-M] 



PROPOSITIONS 9», 93 



199 



Proposition 93. 

//" «k area be contained by a rational straight line and a 
third apotome, the "side" of the area is a second apotome of a 
medial straight line. 

For let the area AB be contained by the rational straight 
line AC and the third apotome AD ; 

I say that the "side" of the area AB is a second apotome of 
a medial straight line. 

For let DG be the annex to AD ; 
therefore AG, GD are rational straight lines commensurable 
in square only, 

and neither of the straight lines AG, GD is commensurable 
in length with the rational straight line AC set out, 
while the square on the whole AG is greater than the square 
on the annex DG by the square on a straight line commen- 



surable with AG. 






[x. Deff. 111. 3] 











c 


} , > 


\ 


K 




R T M 

Since then the square on A G is greater than the square 
on GD by the square on a straight line commensurable 
with AG, 

therefore, if there be applied to AG a parallelogram equal to 
the fourth part of the square on DG and deficient by a square 
figure, it will divide it into commensurable parts. [x. 17] 

Let then DG be bisected at E, 
let there be applied to AG a parallelogram equal to th« 
square on EG and deficient by a square figure, 
and let it be the rectangle AF, EG. 



aoo BOOK X [x. 93 

Let EH, FI, GK be drawn through the points E, F, G 
parallel to AC. 

Therefore AF, FG are commensurable ; 

therefore AI is also commensurable with FK. [vi. 1, x. n] 

And, since AF, FG are commensurable in length, 

therefore AG is also commensurable in length with each of 
the straight lines AF, FG. [x. 15] 

But AG is rational and incommensurable in length 
with AC; 
so that AF, FG are so also. [x. 13] 

Therefore each of the rectangles AI, FK is medial, [x. ai] 

Again, since DE is commensurable in length with EG, 
therefore DG is also commensurable in length with each of 
the straight lines DE, EG. [x. 15} 

But GD is rational and incommensurable in length 
with AC; 

therefore each of the straight lines DE, EG is also rational 
and incommensurable in length with AC; [x. 13] 

therefore each of the rectangles DH, EK is medial. [x. «] 

And, since AG, GD are commensurable in square only ; 

therefore AG is incommensurable in length with GD. 

But AG is commensurable in length with AF, and DG 
with EG ; 

therefore AF is incommensurable in length with EG. [x. 13] 

But, as AF is to EG, so is A I to EK; [vi. 1] 

therefore AI is incommensurable with EK. [x. 1 1] 

Now let the square LMbe, constructed equal to AI, 

and let there be subtracted NO equal to FK and being about 
the same angle with LM; 

therefore LM, NO are about the same diameter, [v<. tt] 

Let PR be their diameter, and let the figure be drawn. 
Now, since the rectangle AF, FG is equal to the square 
on EG, 

therefore, as AF is to EG, so is EG to FG. [vi. 17] 



x. 93] PROPOSITION 93 201 

But, as AF is to EG, so is A I to EK, 
and, as EG is to EG, so is /JTif to FK ; [vi. 1] 

therefore also, as A I is to EK, so is iTA" to EK; [v. 11] 

therefore EK is a mean proportional between AI, FK, 

But J&/7V" is also a mean proportional between the squares 
LM, NO, 

and AI is equal to LM, and FK to jV(? ; 
therefore EK is also equal to MN. 

But jfl/iV is equal to LO, and EK equal to ZW ; 
therefore the whole DK is also equal co the gnomon WW 
and NO. 

But ^ K is also equal to LM, NO ; 
therefore the remainder AB is equal to ST, that is, to the 
square on LN \ 
therefore LN is the "side" of the area AB. 

I say that LN is a second apotome of a medial straight 
line. 

For.since AI,FKvrere proved medial, and are equal to the 
squares on LP, FN, 

therefore each of the squares on LP, PN is also medial ; 
therefore each of the straight lines LP, PN is medial. 

And, since AI is commensurable with FK, [vi. 1, x. n] 
therefore the square on LP is also commensurable with the 
square on PN. 

Again, since A I was proved incommensurable with EK, 
therefore LM is also incommensurable with MN, 
that is, the square on LP with the rectangle LP, PN ; 
so that LP is also incommensurable in length with PN ; 

[vi. i, x. 11] 
therefore LP, PN are medial straight lines commensurable in 
square only. 

I say next that they also contain a medial rectangle. 

For, since EK was proved medial, and is equal to the 
rectangle LP, PN, 

therefore the rectangle LP, PN is also medial, 
so that LP, PN are medial straight lines commensurable in 
square only which contain a medial rectangle. 



*o2 BOOK X [x. 93 

Therefore LN is a second apotome of a medial straight 
line; [x. 7S ] 

and it is the "side" of the area AB. 

Therefore the "side" of the area AB is a second apotome 
of a medial straight line. 

Q. E. D. 
Here we are to find and classify the irrational straight line 

Jp[Ji.p~JA.pjT-k"). 
Following the same method, we put 

v + v=Jk.p 1 

iw=}V(i -* 1 ) ) X } ' 

Next, u, v being found, let 

x* = pu 



f = pv) {2h 



then (x -y) is the square root required and is a second apotome of a medial 
straight line. 

That (x -y) is the square root required and that aft jr are medial areas, so 
that x, y are medial straight lines, is proved exactly as in the last proposition. 

The rectangle xy, being equal to \ Jk . p' «/ 1 - A 1 , is also medial. 

Now, from (i), by x, 17, « " V, 

whence w+ v » u. 

But ( u + *>), of J&-P," i J& ■ P vi- A* ; 

therefore u^^jk.pji — A : , 

and consequently pu v % Ji.p* Ji-k', 

or x* ^ xy, 

whence x vy. 

And, since a « f , pu ^ pv, 

or .r 3 «*j£ 

Thus .v, r are medial straight lines commensurable in square only. 
And .rji is a medial area. 

Therefore (x —y) is a second apotome of a medial straight line. 
Its actual form is found by solving equations (1), (2); 
thus u = HJt.p + KJi.p), 

v = \{Jk.p-Kjk.p), 

and *_^ p yi* (l+ x)_ p y~^ (l _A). 

As explained in the note on x. 56, this is the lesser positive root of the 
equation 

x i -2 l Jk.p>x> + \ i kp' = o. 



x. 94] 



PROPOSITIONS 93, 94 



*e>3 



Proposition 94. 

If an area be contained by a rational straight tine and a 
fourth apotome, the "siaW of the area is minor. 

For let the area AB be contained by the rational straight 
line AC and the fourth apotome AD ; 
I say that the "side" of the area AB is minor. 

For let DG be the annex to AD ; 
therefore AG, GD are rational straight lines commensurable 
in square only, 

AG is commensurable in length with the rational straight line 
AC set out, 

and the square on the whole AG is greater than the square 
on the annex DG by the square on a straight line incommen- 
surable in length with AG, [x. Deff. m. 4] 

A D E F Q 













c 


B 


H 
N 


P 


\ 


t 












s 




t 











u / 


Q,' 








/ 









M 



Since then the square on AG is greater than the square 
on GD by the square on a straight line incommensurable 
in length with AG, 

therefore, if there be applied to AG a parallelogram equal to 
the fourth part of the square on DG and deficient by a square 
figure, it will divide it into incommensurable parts. [x. 18] 

Let then DG be bisected at E, 
let there be applied te AGs. parallelogram equal to the square 
on EG and deficient by a square figure, 
and let it be the rectangle AF, FG ; 
therefore AF is incommensurable in length with FG. 



ao4 BOOK X [x. 94 

Let EH, FI, GK be drawn through E, F, G parallel to 
AC, BD. 

Since then AG is rational and commensurable in length 
with AC, 
therefore the whole AK is rational. [x. 19] 

Again, since DG is incommensurable in length with AC, 
and both are rational, 
therefore DK is medial. [x. 21] 

Again, since AF is incommensurable in length with FG, 
therefore A I is also incommensurable with FK. [vi. 1, x. n] 

Now let the square LM be constructed equal to A I, 
and let there be subtracted NO equal to FK and about the 
same angle, the angle LPM. 

Therefore the squares LM, NO are about the same 
diameter. [vi. »6] 

Let PR be their diameter, and let the figure be drawn. 

Since then the rectangle AF, FG is equal to the square 
on EG, 

therefore, proportionally, as AF is to EG, so is EG to FG. 

[vi. 17] 

But, as AF is to EG, so is AI to EK, 
and, as EG is to FG, so is EK to FK ; [vi. 1] 

therefore EK is a mean proportional between A I, FK. [v. n] 

But MN is also a mean proportional between the squares 
LM, NO, 

and AI is equal to LM, and FK to NO ; 
therefore EK is also equal to MN. 

But DH is equal to EK, and LO is equal to ■#/A r ; 

therefore the whole DK is equal to the gnomon UVW 
and JVa 

Since, then, the whole A K is equal to the squares 
LM, NO, 

and, in these, DK is equal to the gnomon UVW and the 
square NO, 

therefore the remainder AB is equal to ST, that is, to the 
square on LN ; 
therefore LN is the "side" of the area AB. 



x. 94] PROPOSITION 94 105 

I say that LN is the irrational straight line called minor. 

For, since AK is rational and is equal to the squares on 
LP, PN, 
therefore the sum of the squares on LP, PN is rational. 

Again, since DK is medial, 
and DK is equal to twice the rectangle LP, PN, 
therefore twice the rectangle LP, PN is medial. 

And, since A J was proved incommensurable with FK, 

therefore the square on LP is also incommensurable with the 
square on PN. 

Therefore LP, PN axe, straight lines incommensurable in 
square which make the sum of the squares on them rational, 
but twice the rectangle contained by them medial. 

Therefore LN'\s the irrational straight line called minor; 

[x. 76] 
and it is the "side" of the area AB. 

Therefore the "side" of the area AB is minor. 

Q. E, D, 
We have here to find flhd classify the straight line 



equations 
1 = kp \ 

* 1 + X ) 



As usual, we find u, v from the equations 

u + v = kp 



uv- 
and then, giving u, v their values, we put 

*'.*} » 

Then (x— y) is the required square root. 

This is proved in the same way as before, and, as before, it is proved that 

Now, from (r), by x. 18, u •*, v; 

therefore pu w pv, 

Or &"?, 

so that x, y are incommensurable in square. 

And .** +y*, or p (a + v), is a rational area {kp 1 ). 

it 
But ixy = 1 , which is a medial area. 

Hence [x. 76] (x~y) is the irrational straight line called minor. 



ao« BOOR X [x. 94, 9$ 

Algebraical solution gives 

whence .-^p^f. + s/^-V^' " \/rh)' 

As explained in the note on x. 57, this is the lesser positive root of the 
equation 

X> - 2 V . X* + -^-r k*f? - O. 



^- 






1 + A 



Proposition 95. 

If an area be contained by a rational straight line and a 
fifth apotome, the "side" of the area is a straight line which 
produces with a rational area a medial whole. 

For let the area AB be contained by the rational straight 
line AC and the fifth apotome AD\ 

I say that the " side " of the area AB is a straight line which 
produces with a rational area a medial whole. 

For let DG be the annex to AD ; 
therefore AG, GD are rational straight lines commensurable 
in square only, 











c 




3 H 


1 K 

p 




L 
S 






u / 


y 


1 


1 


1 


r 


*, 


1 



the annex GD is commensurable in length with the rational 

straight line AC set out, 

and the square on the whole AG is greater than the square 



x. 95] PROPOSITIONS 94, 95 207 

on the annex DG by the square on a straight line incommen- 
surable with AG. [x. Deff. 111. 5] 

Therefore, if there be applied to AG a parallelogram 
equal to the fourth part of the square on DG and deficient 
by a square figure, it will divide it into incommensurable 
parts. [x. 18] 

Let then DG be bisected at the point E, 

let there be applied to AG a parallelogram equal to the 
square on EG and deficient by a square figure, and let it be 
the rectangle AF, FG ; 
therefore AF is incommensurable in length with FG. 

Now, since AG is incommensurable in length with CA, 
and both are rational, 

therefore AK is medial. [x. 21] 

Again, since DG is rational and commensurable in length 
with AC, 

DK\s rational. [x. 19] 

Now let the square LM be constructed equal to A/, and 
let the square NO equal to FK and about the same angle, the 
angle LPM, be subtracted ; 

therefore the squares LM, NO are about the same diameter. 

[vi. 26] 

Let PR be their diameter, and let the figure be drawn. 

Similarly then we can prove that LN is the "side" of the 
area AB. 

I say that LN is the straight line which produces with a 
rational area a medial whole. 

For, since AK was proved medial and is equal to the 
squares on LP, PN, 

therefore the sum of the squares on LP, PN is medial. 

Again, since DK is rational and is equal to twice the 
rectangle LP, PN, 
the latter is itself also rational. 

And, since AI is incommensurable with FK, 
therefore the square on LP is also incommensurable with the 
square on PN ; 
therefore LP, PN are straight lines incommensurable in 



so8 BOOK X [x. 95 

square which make the sum of the squares on them medial 
hut twice the rectangle contained by them rational. 

Therefore the remainder LN is the irrational straight line 
called that which produces with a rational area a medial 
whole ; [x. 77] 

and it is the " side " of the area AB. 

Therefore the "side" of the area AB is a straight line 
which produces with a rational area a medial whole. 

Q. E. D. 
Here the problem is to find and classify 



■Jp(ipJTTk-kp). 
As usual, we put 

u + v = &p<Ji+k) . . 

uv*lt>? I *** 

and, v, v being found, we take 

* = **\ <«). 

Then {x -y) so found is our required square root. 
This fact is proved as before, and, as before, we see that 

Now from (1), by x. -8, v ~ v, 

whence P« ^ pv, 

or ^u/, 

and .v, y are incommensurable in square. 

Next (x* + y*) = p (a + v) = kf? J i + K, which is a medial area. 

And ixy = ip\ which is a rational area. 

Hence (x-y) is the "side" of a medial, minus a rational, area. [x. 77] 

Algebraical solution gives 

2 

v~ — 
and therefore 



f (Vn-X-VA), 



which is, as explained in the note to x. 58, the lesser positive root of the 
equation 

x* - ikp* V77A. . x* + X^p* = o. 



x. 96] 



PROPOSITIONS 95, 96 



209 



Proposition 96. 

If an area be contained by a rational straight line and a 
sixth apotome, the "side" of the area is a straight line which 
produces with a medial area a medial whole. 

For let the area AB be contained by the rational straight 
line AC and the sixth apotome AD ; 

I say that the "side" of the area AB is a straight line which 
produces with a medial area a medial whole. 

A d e f a 




H I K 









For let DG be the annex to AD ; 

therefore AG, GD are rational straight lines commensurable 
in square only, 

neither of them is commensurable in length with the rational 
straight line AC set out, 

and the square on the whole A G is greater than the square 
on the annex DG by the square on a straight line incommen- 
surable in length with AG. [x. Deff. in. 6] 

Since then the square on AG is greater than the square 
on GD by the square on a straight line incommensurable in 
length with AG, 

therefore, if there be applied to AG a parallelogram equal to 
the fourth part of the square on DG and deficient by a square 
figure, it will divide it into incommensurable parts. [x. 18; 

Let then DG be bisected at E, 

let there be applied to AG a parallelogram equal to the square 



no BOOK X [x. 96 

on EG and deficient by a square figure, and let it be the 

rectangle AF, FG ; 

therefore AF is incommensurable in length with FG. 

But, as AFis to FG, so is AI to FK; [vi.i] 

therefore A I is incommensurable with FK. [x. u] 

And, sincere AC are rational straight lines commensur- 
able in square only, 
AK is medial. [x. 21] 

Again, since AC, DG are rational straight lines and 
incommensurable in length, 
DK is also medial. [x. 21] 

Now, since AG, GD are commensurable in square only, 
therefore AG is incommensurable in length with GD. 

But, as A G is to GD, so is AK to KD ; [vi. 1] 

therefore A K is incommensurable with KD. [x. n] 

Now let the square LM be constructed equal to ^4/, 
and let NO equal to AA", and about the same angle, be 
subtracted ; 

therefore the squares LM, NO are about the same diameter. 

[vi. 26] 

Let PR be their diameter, and let the figure be drawn. 

Then in manner similar to the above we can prove that 
LN is the " side " of the area AB. 

I say that LN is a straight line which produces with a 
medial area a medial whole. 

For, since AK was proved medial and is equal to the 
squares on LP, PN, 
therefore the sum of the squares on LP, PN is medial. 

Again, since DK was proved medial and is equal to twice 
the rectangle LP, PN, 
twice the rectangle LP, PN is also medial. 

And, since AK was proved incommensurable with DK, 
the squares on LP, PN are also incommensurable with twice 
the rectangle LP, PN. 

And, since AT is incommensurable with FK, 
therefore the square on LP is also incommensurable with the 
square 011 PN; 



X. 96] PROPOSITION 96 an 

therefore LP, PN are straight lines incommensurable in 
square which make the sum of the squares on them medial, 
twice the rectangle contained by them medial, and further the 
squares on them incommensurable with twice the rectangle 
contained by them. 

Therefore LN is the irrational straight line called that 
which produces with a medial area a medial whole; [x. 78] 

and it is the " side " of the area AB. 

Therefore the "side" of the area is a straight line which 
produces with a medial area a medial whole. 



Q. E. D. 



We have to find and classify 



>Jp{^.p-jK.p). 
Put, as usual, 

w=£V I [ A 

and, u, v being thus found, let 

J = '"} <*). 

Then, as before, (x -y) is the square root required. 
For, from (1), by x. iS, w ~ v, 

whence pu \j pv, 

or ^wy, 

and x, y are incommensurable in square. 

Next, Jt 1 +y* = p (» +■ i>) = JA . p", which is a media} area. 
Also 2xy = JX. . p', which is again a medial area. 
Lastly, Jk . ft Jk . p are by hypothesis 1^, so that 

Jk.pv jk.p, 

whence Jk . p' v J\, p*, 

or (x'+y r ) -j 2xy. 

Thus (x-y) is the "side" of a medial, minus a medial, area [x. 78]. 

Algebraical solution gives 

whence * -J- = pVi (V* + -J*^K) - p %/j (V* " V*^A). 

This, as explained in the note on x. 59, is the lesser positive root of the 
equation 

x l -iJk.p'x*+(i-K)p t = o. 



212 



BOOK. X 



[x.97 



Proposition 97. 

The square on an apotome applied to a rational straight 
line produces as breadth a first apotome. 

Let AB be an apotome, and CD rational, 
and to CD let there be applied CE equal to the square on 
AB and producing CF as breadth ; 
I say that CF is a first apotome. 

A B Q 

C F N K M 



For let BG be the annex to AB ; 
therefore AG, GB are rational straight lines commensurable 
in square only. [x. 73] 

To CD let there be applied CH equal to the square on 
AG, and KL equal to the square on BG. 

Therefore the whole CL is equal to the squares on AG, GB, 
and, in these, CE is equal to the square on AB; 
therefore the remainder FL is equal to twice the rectangle 
AG, GB. [». 7] 

Let FM be bisected at the point N, 
and let NO be drawn through TV parallel to CD ; 
therefore each of the rectangles FO, LN is equal to the 
rectangle AG, GB. 

Now, since the squares on AG, GB are rational, 
and DM is equal to the sq uares on A G, GB, . 
therefore DM is rational. 

And it has been applied to the rational straight line CD, 
producing CM as breadth ; 

therefore CM is rational and commensurable in length with 
CD. [x. ao] 

Again, since twice the rectangle AG, GB is medial, and 
FL is equal to twice the rectangle AG, GB, 
therefore FL is medial. 



x. 97 ] PROPOSITION 97 113 

And it is applied to the rational straight line CD, producing 
FM as breadth ; 

therefore FM is rational and incommensurable in length with 
CD. [x. 22] 

And, since the squares on AG, GB are rational, 

while twice the rectangle AG, GB is medial, 

therefore the squares on AG, GB are incommensurable with 
twice the rectangle AG, GB. 

And CL is equal to the squares on AG, GB, 
and FL to twice the rectangle A G, GB ; 
therefore DM is incommensurable with FL. 

But, as DM is to FL, so is CM to FM; [vi. 1] 

therefore CM is incommensurable in length with FM. [x. n] 

And both are rational ; 

therefore CM, MF are rational straight lines commensurable 
in square only ; 

therefore CF is an apotome. [x. 73] 

I say next that it is also a first apotome. 
For, since the rectangle AG, GB is a mean proportional 
between the squares on AG, GB, 

and CH is equal to the square on AG, 

KL equal to the square on BG, 

and NL equal to the rectangle AG, GB, 

therefore NL is also a mean proportional between CH, KL ; 

therefore, as CH is to NL, so is NL to KL. 

But, as CH is to NL, so is CK to NM, 

and, as NL is to KL, so is NM to KM; [vi. 1] 

therefore the rectangle CK, KM is equal to the square on 
NM [vi. 1 7], that is, to the fourth part of the square on FM. 

And, since the square on AG is commensurable with the 
square on GB, 
CH is also commensurable with KL. 

But, as CH is to KL, so is CK to KM ; [vi. 1] 

therefore CK is commensurable with KM. [*■ "] 

Since then CM, MF are two unequal straight lines, 



ii4 BOOK X [x. 97 

and to CM there has been applied the rectangle CK, KM 

equal to the fourth part of the square on FM and deficient by 

a square figure, 

while CK is commensurable with KM, 

therefore the square on CM is greater than the square on MF 

by the square on a straight line commensurable in length 

with CM. Ex. 17] 

And CM is commensurable in length with the rational 
straight line CD set out ; 
therefore CF is a first apotome. [x. Deff. in. 1] 

Therefore etc. 

Q. E. D. 

Here begins the hexad of propositions solving the problems which are the 
converse of those in the hexad just concluded- Props. 97 to ioj correspond 
of course to Props. 60 to 65 relating to the binomials etc. 

We have in x. 97 to prove that, (p — Jk . p) being an apotome, 

{p~Jk.pf 
* 
is a first apotome, and we have to find it geometrically. 
Euclid's procedure may be represented thus. 
Take x, y, z such that 

«-• J 

<y = V } (')■ 

<r . n = 2 *jk . p* ' 

Thus (x + y)-2sJ- p -* /i - ( ' )t , 

and we have to prove that (x+y) - 22 is a first apotome. 

(a) Now jj 5 + ifj\ or <r (x +y), is rational ; 

therefore (x+y) is rational and *a (2)- 

And 2 Jk . p*, or <r . 2 s, is medial : 
therefore 22 is rational and ^ <r (3). 

But, cr (x 4-y) being rational, and a . %z metlial, 
<r(x + y) w<r. it, 
whence (* + „v) <* "• 

Therefore, since (x +y), 2Z are both rational [(2), (3)], 
(x +J>), 22 are rational and <■*- (4). 

Hence (x +y)- 22 is an apotome. 

(ft) Since Jk . p* is a mean proportional between p s , ip*, 
o-a is a mean proportional between ax, ay [by (1)]. 

That is, . ax : an = az : ay, 
or x : z = z • y, 
and *y = *, or i(a*)' (5). 



PROPOSITIONS 97, 98 



x. 97, 98] 

And, since p* « kp 1 , <rx n ory, 
or j; " j 

Hence [{5), (6)], by X. 17, 

■J{x + >)' - (is)' " (x +Jf). 

And [(4)] {* +ji), iz are rational and '«-, 
while [(2)] {*+.)') « a ; 
therefore (.r +J>) — 2Z is ajf.r.ir' apotgme. 

The actual value of (jt + y) - as is of course 



aiS 



...(6). 



Proposition 98. 

The square on a first apotome of a medial straight line 
applied to a rational straight line produces as breadth a second 
apotome. 

Let AB be a first apotome of a medial straight line and 
CD a rational straight line, 

and to CD let there be applied CM equal to the square on 
AB, producing CF as breadth ; 
I say that CF is a second apotome. 

For let BG be the annex to AB ;. 
therefore AG, GB are medial straight lines commensurable in 
square only which contain a rational rectangle. [x. 74] 

A 6 G 



F 


< M 










D E 


: c 


3 


4 L 



To CD let there be applied C.H equal to the square on 
AG, producing CK as breadth, and KL equal to the square 
on GB, producing KM as breadth ; 

therefore the whole CL is equal to the squares on A G, GB ; 
therefore CL is also medial. [".15 and 23, Por.] 

And it is applied to the rational straight line CD, pro- 
ducing CM as breadth ; 

therefore CM is rational and incommensurable in length with 
CD. [x. 22] 



2t6 BOOK X [x. 98 

Now, since CL is equal to the squares on AG, GB, 
and, in these, the square on AB is equal to CE, 
therefore the remainder, twice the rectangle A G, GB, is equal 
to FL. [n. 7] 

But twice the rectangle AG, GB is rational ; 
therefore FL is rational. 

And it is applied to the rational straight line FE, producing 
FM as breadth ; 

therefore FM is also rational and commensurable in length 
with CD. [x. 20] 

Now, since the sum of the squares on AG, GB, that is, 
CL, is medial, while twice the rectangle AG, GB, that is, FL, 
is rational, 
therefore CL is incommensurable with FL. 

But, as CL is to FL, so is CM to FM; [vi. 1] 

therefore CM is incommensurable in length with FM. [x. n] 

And both are rational ; 
therefore CM, MF are rational straight lines commensurable 
in square only ; 

therefore CF is an apotome. [x. 73] 

I say next that it is also a second apotome. 

For let FM be bisected at N, 
and let NO be drawn through A^ parallel to CD ; 
therefore each of the rectangles FO, NL is equal to the 
rectangle A G, GB. 

Now, since the rectangle AG, GB is a mean proportional 
between the squares on AG, GB, 
and the square on AG is equal to CH, 
the rectangle AG, GB to NL, 
and the square on BG to KL, 

therefore NL is also a mean proportional between CH, KL; 
therefore, as CH is to NL, so is NL to KL. 

But, as CH is to NL, so is CK to NM, 
and, as NL is to KL, so is NM to MK; [vi. 1] 

therefore, as CK is to NM, so is NM to KM '; [v. 11] 

therefore the rectangle CK, KM is equal to the square on 
NM [vi. 17], that is, to the fourth part of the square on FM, 



x. 9 8] PROPOSITION 98 *i7 

Since then CM, MF are two unequal straight lines, and 
the rectangle CK, KM equal to the fourth part of the square 
on MF and deficient by a square figure has been applied to 
the greater, CM, and divides it into commensurable parts, 
therefore the square on CM is greater than the square on MF 
by the square on a straight line commensurable in length with 
CM. [x. .7] 

And the annex FM is commensurable in length with the 
rational straight line CD set out ; 
therefore CF is a second apotome. [x, Deff. 111. a] 

Therefore etc. 

Q. E. D. 

In this case we have to find and classify 

Take x, y, z such that 

ax = $p' \ 

■?=>**/ I i*y 

<r , tz = Jjfp* I 

(a) Now £y, 4V are medial areas ; 

therefore <r {x + y) is medial, 

whence &+y) is rational and ^ <r , (2). 

But zip 8 , and therefore o- . zsr, is rational, 
whence iz is rational and *• <r (3). 

And, <r(x + y) being medial, and er . iz rational, 
a ( x + y) vj <r . 2Z, 
or {jc +_y) v zz. 

Hence (jc + _y), za are rational straight lines com mens ura"ble in square only, 
and therefore (x + y)-nz is an apvtofne, 

{&) We prove, as before, that 

«M4Ktf (4). 

Also k?f? f* k'p', or ax « try, 
so that :r"jr (5). 

[This step is omitted in P, and Heiberg accordingly brackets it. The 
result is, however, assumed.] 
Therefore [(4), (5)], by x, 17, 



■A* +y)' - (2Z} 1 « (* +y). 
And 22 « <r. 
Therefore (x +y) - iz is a second apotome. 

o> 
Obviously (x +_y) - 2% - - \JA (i + k) - 2k). 



2i8 



BOOK X 



[x. 99 



Proposition 99. 

The square on a second apotome of a medial straight line 
applied to a rational straight line produces as breadth a third 

apotome. 

Let AB be a second apotome of a medial straight line, 
and CD rational, 

and to CD let there be applied CE equal to the square on 
AB, producing CF as breadth ; 
I say that CF is a third apotome. 



M 



For let BG be the annex to AB ; 
therefore AG, GB are medial straight lines commensurable 
in square only which contain a medial rectangle. [x. 75] 

Let CH equal to the square on AG be applied to CD, 
producing CK as breadth, 

and let KL equal to the square on BG be applied to KH t 
producing KM as breadth ; 

therefore the whole CL is equal to the squares on A G, GB ; 
therefore CL is also medial- [x. 15 and 23, Por.] 

And it is applied to the rational straight line CD, producing 
CM as breadth ; 

therefore CM is rational and incommensurable in length with 
CD. [x. 22] 

Now, since the whole CL is equal to the squares on AG, 
GB, and, in these, CE is equal to the square on AB, 
therefore the remainder LF is equal to twice the rectangle 
AG,GB. [h.7] 

Let then FM be bisected at the point N, 
and let NO be drawn parallel to CD ; 

therefore each of the rectangles FO, NL is equal to the rect- 
angle AG, GB. 



x. 99] PROPOSITION 99 »ig 

But the rectangle AG, GB is medial ; 
therefore FL is also medial. 

And it is applied to the rational straight line EF, producing 
FM as breadth ; 

therefore FM is also rational and incommensurable in length 
with CD. [x. **] 

And, since AG, GB are commensurable in square only, 
therefore AG is incommensurable in length with GB; 
therefore the square on A G is also incommensurable with the 
rectangle AG, GB. [vi. i, x. n] 

But the squares on AG, GB are commensurable with the 
square on AG, 

and twice the rectangle A G, GB with the rectangle A G, GB ; 
therefore the squares on AG, GB are incommensurable with 
twice the rectangle AG, GB. [x. 13] 

But CL is equal to the squares on AG, GB, 
and FL is equal to twice the rectangle AG, GB ; 
therefore CL is also incommensurable with FL. 

But, as CL is to FL, so is CM to FM; [vi. 1] 

therefore CM is incommensurable in length with FM. [x. 11] 

And both are rational ; 

therefore CM, MFare rational straight lines commensurable 
in square only ; 

therefore CF is an apotome. [x, 73] 

I say next that It is also a third apotome. 
For, since the square on AG is commensurable with the 
square on GB, 

therefore CH is also commensurable with KL t 
so that CK is also commensurable with KM. [vi. 1, x. it] 

And, since the rectangle AG, GB Is a mean proportional 
between the squares on AG, GB, 

and CH is equal to the square on A G, 

KL equal to the square on GB, 

and NL equal to the rectangle AG, GB, 

therefore NL is also a mean proportional between CH, KL ; 

therefore, as CH is to NL, so is NL to KL. 



220 BOOK X [x. 99 

But, as CH is to NL, so is CK to NM, 
and, as NL is to KL, so is JVJSf to KM ; [vj. i] 

therefore, as CK is to MN, so is ^/.A/ to KM; [v. n] 

therefore the rectangle CK, KM is equal to [the square on 
MN, that is, to] the fourth part of the square on FM. 

Since then CM, MF are two unequal straight lines, and 
a parallelogram equal to the fourth part of the square on FM 
and deficient by a square figure has been applied to CM, and 
divides it into commensurable parts, 

therefore the square on CM is greater than the square on 
MF by the square on a straight line commensurable with 
CM. [x. 17] 

And neither of the straight lines CM, MF is commensur- 
able in length with the rational straight line CD set out ; 
therefore CF is a third apotome. [x. Deff. 111. 3] 

Therefore etc. 



Q. E. D. 



We have to find and classify 






Take x, y, 2 such that 



Ji.p- 



X 

J* 









*~W 



(a) Then tr(x + y) is a medial area, 

whence (x +y) is rational and ^ <r (1). 

Also a . 2Z is medial, 
whence az is rational and w a (z). 

Again ^~^r> 

whence Jk .(? <-> J\ . p*. 

And Jk.p^^Jk.ff + ^f?), 

while ^/K . p' « 2 JK . (? ; 

therefore f Jk . p* + —p, p* J ^ i^k.p\ 

or o- (jr +_y) s/ <r . 2z, 

and (* + /) ** « (3) 




























x. 99, ioo] 



PROPOSITIONS 99, ioo 



Thus f_{i), (2), (3}] (x +y), 23 are rational and "-, 
so that (x +y) — 2Z is an apotome. 

{£) trx f try, so that x ^ y. 

And, as before, xy = \{ts)\ 

Therefore [x. 17] ^(x+yf-iiz)' « (*• + y). 
And neither (x +y) nor iz is « <r. 
Therefore {* + _y) - 22 is a Ai/rtf apotome. 
It is of course equal to 



221 






Proposition ioo. 

The square on a minor straight line applied to a rational 
straight line produces as breadth a fourth apotome. 

Let AB be a minor and CD a rational straight line, and 
to the rational straight line CD let CE be applied equal to the 
square on AB and producing CF as breadth ; 
I say that CF'xs, a fourth apotome. 



-S- 



M 



For let BG be the annex to AB ; 
therefore AG, GB are straight lines incommensurable in 
square which make the sum of the squares on AG, GB 
rational, but twice the rectangle AG, GB medial. [x. 76] 

To CD let there be applied CH equal to the square on 
A G and producing CK as breadth, 

and KL equal to the square on BG, producing KM as breadth ; 
therefore the whole CL is equal to the squares on AG, GB. 

And the sum of the squares on AG, GB is rational ; 
therefore CL is also rational. 

And it is applied to the rational straight line CD, producing 
CM as breadth ; 

therefore CM is also rational and commensurable in length 
with CD, [x. to] 



a« BOOK. X [x. 100 

AikI, since the whole CL is equal to the squares on AG, 
GB, and, in these, CM is equal to the square on AB, 

therefore the remainder FL is equal to twice the rectangle 
AG, GB. [u. 7] 

Let then FM be bisected at the point N, 

and let NO be drawn through N parallel to either of the 
straight lines CD, ML ; 

therefore each of the rectangles FO, NL is equal to the rect- 
angle AG, GB. 

And, since twice the rectangle AG, GB is medial and is 
equal to FL, 

therefore FL is also medial. 

And it is applied to the rational straight line FE, producing 
FM as breadth ; 

therefore FM is rational and incommensurable in length with 
CD. [x. »] 

And, since the sum of the squares on AG. GB is rational, 
while twice the rectangle AG, GB is medial, 
the squares on AG, GB are incommensurable with twice the 
rectangle AG, GB. 

But CL is equal to the squares on AG, GB, 
and FL equal to twice the rectangle AG, GB ; 
therefore CL is incommensurable with FL. 

But, as CL is to FL, so is CM to MF; [vi. i] 

therefore CM is incommensurable in length with MF. [x. n] 

And both are rational ; 
therefore CM, MF are rational straight lines commensurable 
in square only ; 
therefore CF is an apotome. [x. 73] 

I say that it is also a fourth apotome. 

For, since AG, GB are incommensurable in square, 

therefore the square on AG is also incommensurable with the 
square on GB, 

And CH is equal to the square on A G, 
and KL equal to the square on GB \ 
therefore CH is incommensurable with KL. 



x. ioo] PROPOSITION 100 223 

But, as CH is to KL, so is CK to KM; [vi. 1} 

therefore CA" is incommensurable in length with KM. [x. n] 

And, since the rectangle AG, GB is a mean proportional 
between the squares on AG, GB, 
and the square on AG is equal to CH, 
the square on GB to KL, 
and the rectangle AG, GB to NL, 
therefore NL is a mean proportional between CH, KL j 
therefore, as £7/ is to NL, so is jVX to KL. 

But, as C7/ is to NL, so is CK to AW, 

and, as NL is to ATZ, so is NM to AW; [vi i] 

therefore, as CK is to MN, so is iJ/A'' to KM; [v, u] 

therefore the rectangle CK, KM is equal to the square on 
MN [vi. i 7], that is, to the fourth part of the square on FM. 

Since then CM, MF are two unequal straight lines, and 
the rectangle CK, KM equal to the fourth part of the square 
on MF and deficient by a square figure has been applied to 
CM and divides it into incommensurable parts, 

therefore the square on CM is greater than the square on 
MF by the square on a straight line incommensurable with 
CM. [x. 18] 

And the whole CM is commensurable in length with the 
rational straight line CD set out ; 

therefore CF is a fourth apotome. [x. Deff. m. 4] 

Therefore etc. 

Q. E. D. 

We have to find and classify 
We will call this, for brevity, 



■ 



Take x, y, z such that 









ty = & }, 
a . 2* = 2OTJ J 

where it has to be remembered that tt\ 1? are incommensurable, («* + »/") if 
rational, and tuv medial. 



214 



BOOK X 



[x. 100, IOI 



It follows that tr(x+y) is rational and o- . 22 medial, 

so that (x -t-y) is rational and « o- 

while 2£ is rational and u<r 

and a(x + y) ut.jj, 

so that {x+y)^zz 

Thus [(i), (z), (3)] (# +>), 2J are rational and >*-, 
so that (,* +^>) — 2« is an apotome. 

Next, since »* ^ i?, 

<r jc -j <jy, 
or * „ ji. 

And it is proved, as usual, that 

xy = s t = i(tz) t . 

Therefore [x. 18] *J(x +;•)•-"( izf ~ (x+y). 

But (x+y) « <r, 
therefore jr+_) p — 21 is a. fourth apotomt. 

P* 



■(0. 

(3). 



Its value is of course 






Proposition 101. 

7"^« square on the straight line which produces with a 
rational area a medial whole, if applied to a rational straight 
line, produces as breadth a fifth apotome. 

Let AB be the straight line which produces with a 
rational area a medial whole, and CD a rational straight line, 
rtnd to CD let C£ be applied equal to the square on AB and 
producing CF as breadth ; 
I say that CF is a fifth apotome. 

a no 



M 



d e 

For let BG be the annex to AB ; 
therefore AG, GB are straight lines incommensurable in 
square which make the sum of the squares on them medial 
but twice the rectangle contained by them rational. [x. 77] 

To CD let there be applied CM equal to the square on 
A G, and KL equal to the square on GB ; 
therefore the whole CL is equal to the squares on AG, GB. 



x. ioi] PROPOSITIONS ioo, 101 125 

But the sum of the squares on AG, GB together is 
medial ; 

therefore CL is medial. 

And it is applied to the rational straight line CD, producing 
CM as breadth ; 

therefore CM is rational and incommensurable with CD. [x. 22] 
And, since the whole CL is equal to the squares on AG, GB, 
and, in these, CE is equal to the square on AB, 
therefore the remainder FL is equal to twice the rectangle 
AG, GB. [11. 7] 

Let then FM be bisected at N, 

and through N let NO be drawn parallel to either of the 
straight lines CD, ML ; 

therefore each of the rectangles FO, NL is equal to the rect- 
angle AG, GB'. 

And, since twice the rectangle AG, GB is rational and 
equal to FL, 
therefore FL is rational. 

And it is applied to the rational straight line EF, producing 
FM as breadth ; 

therefore FM is rational and commensurable in length with 
CD. [x. ao] 

Now, since CL is medial, and FL rational, 
therefore CL is incommensurable with FL. 

But, as CL is to FL, so is CM to MF; [vi. 1] 

therefore CM is incommensurable in length with MF. [x. n] 

And both are rational ; 
therefore CM, MF are rational straight lines commensurable 
in square only ; 
therefore CF is an apotome. [x. 73] 

I say next that it is also a fifth apotome. 

For we can prove similarly that the rectangle C K, KM 
is equal to the square on NM, that is, to the fourth part of the 
square on FM. 

And, since the square on AG is incommensurable with the 
square on GB, 



226 BOOK X [x. ioi 

while the square on AG is equal to CH, 

and the square on GB to KL, 

therefore CH is incommensurable with KL . 

But, as CH is to KL, so is CK to KM ; [vi. i] 

therefore CK 'is incommensurable in length with KM. [x. "] 

Since then CM, MFare two unequal straight lines, 
and a parallelogram equal to the fourth part of the square 
on FM and deficient by a square figure has been applied to 
CM, and divides it into incommensurable parts, 
therefore the square on CM is greater than the square on 
MF by the square on a straight line incommensurable with 
CM. [x. i8j 

And the annex FM is commensurable with the rational 
straight line CD set out ; 
therefore CF is a fifth apotome. |x. Deff, hi. 5) 

Q. E. D. 

We have to Pod and classify 



«r Wa<t+#) 



*/ah +jP1 1 



Call this - (« - r)*> and take a-, _v, z such that 

<rx = u' 1 

<TV = *» . 

cr. M- i«p ) 

In this case « s . #* are incommensurable, (a* + 1?) is a medial area and iuv 
a rational area. 

Since it{x + j>) is medial and <r . 2z rational, 
(x +y) is rational and w c\ 
2* is rational and A tr, 
while (*+>) - ** 

It follows that (# +_>>), 2% are rational and «-, 
so that {*+.?) - ** is an apotome. 

Again, as before, xy = z a = \ (mf, 

and, since «' v **, 03 ^ try, 

or x v y. 

Hence [x. 18]- ^+>')'-(a«)' « (*+>). 

And 32 " <r. 

Therefore {■* +.y) - 22 is a fifth apotemt. 

It is of course equal to 

f. - 



£/ _* L_V 



X. I01J 



PROPOSITIONS 101, ioj 



1*7 






Proposition 102. 



The square on the straight line which produces with a 
medial area a medial whole, if applied to a rational straight 
line, produces as breadth a sixth apotome. 

Let AB be the straight line which produces with a medial 
area a medial whole, and CD a rational straight line, 

and to CD let CE be applied equal to the square on AB and 
producing CF as breadth ; 

I say that CF is a sixth apotome. 



M 






For let BG be the annex to AB ; 

therefore AG, GB are straight lines incommensurable in 
square which make the sum of the squares on them medial, 
twice the rectangle AG, GB medial, and the squares on AG, 
GB incommensurable with twice the rectangle AG, GB. [x. j8] 

Now to CD let there be applied CH equal to the square 
on AG and producing CK as breadth, 

and KL equal to the square on BG ; 

therefore the whole CL is equal to the squares on AG, GB; 

therefore CL is also medial. 

And it is applied to the rational straight line CD, produc- 
ing CM as breadth ; 

therefore CM is rational and incommensurable in length 
with CD. [x. »a) 

Since now CL is equal to the squares on AG, GB, 
and, in these, CF is equal to the square on AB, 
therefore the remainder FL is equal to twice the rectangle 
AG, GB. [11. 7] 

And twice the rectangle AG, GB is medial ; 
therefore FL is also medial. 



zzS BOOK X [x. 102 

And it is applied to the rational straight line FE, pro- 
ducing FM as breadth ; 

therefore FM is rational and incommensurable in length 
with CD, [x. 21] 

And, since the squares on AG, GB are incommensurable 
with twice the rectangle AG, GB, 
and CL is equal to the squares on AG, GB, 
and FL equal to twice the rectangle AG, GB, 
therefore CL is incommensurable with FL. 

But, as CL is to FL, so is CM to MF; [n. 1] 

therefore CM is incommensurable in length with MF, [x. n] 

And both are rational. 

Therefore CM, MF are rational straight lines commen- 
surable in square only ; 
therefore CF is an apotome. [x. 73] 

I say next that it is also a sixth apotome. 

For, since FL is equal to twice the rectangle AG, GB, 
let FM be bisected at N, 

and let NO be drawn through N parallel to CD ; 
therefore each of the rectangles FO, NL is equal to the rect- 
angle AG, GB. 

And, since AG, GB are incommensurable in square, 

therefore the square on AG is incommensurable with the 
square on GB. 

But CH is equal to the square on AG, 
and KL is equal to the square on GB ; 
therefore CH is incommensurable with KL. 

But, as CH is to KL, so is CK to KM ; [vi, 1] 

therefore CK is incommensurable with KM. [x. n] 

And, since the rectangle AG, GB is a mean proportional 
between the squares on AG, GB, 

and CH is equal to the square on AG, 

KL equal to the square on GB, 

and NL equal to the rectangle AG, GB, 

therefore NL is also a mean proportional between CH, KL ; 

therefore, as CH is to NL, so is NL to KL. 



x. ioi, 103] PROPOSITIONS 102, 103 a 29 

And for the same reason as before the square on CM is 
greater than the square on MF by the square on a straight 
line incommensurable with CM. [x. 18] 

And neither of them is commensurable with the rational 
straight line CD set out ; 
therefore CF is a sixth apotome. [x. Deff. hi. 6] 

Q. E. D. 

We have to find and classify 

i/ei\/ I + _jt f$ A_ * \\ 

rWiV VJ+s T^v TxWj 



Call this - (u - v)', and put 



<r* = **, 












ff . 2S = XUV. 

Here «*, w° are incommensurable, 
{u* + s*), 2uv are both medial areas, 
and (« a + »*) ^ iuv. 

Since o- (a+^), o- . 22 are medial and incommensurable, 
{x + y) is rational and ^ o-, 
zz is rational and w <r, 
and (•*+>) « as. 

Hence {.* + _?), az are rational and "-, 
so that (x +y) — 22 is an apotome. 

Again, since a', p*, or cms, ory, are incommensurable, 

And, as before, xy ■ *" = i (m)'. 

Therefore [x. t8] V(i +.?)* - (22)' « (*+.>'). 
And neither (x+y) nor 22 is « z; 
therefore (a +/) — 2* is a (£«C4 apotome. 

It is of course - [ JK — , ■ ■) . 

Proposition 103. 

A straight line commensurable in length with an apotome 
is an apotome and the same in order. 

Let AB be an apotome, 

and let CD be commensurable in a j E 

length with AS ; Q p F 

I say that CD is also an apotome and 
the same in order with AB. 



J30 BOOK X [x. 103 

For, since AB is an apotome, let BE be the annex to it ; 

therefore AE, EB are rational straight lines commensurable 
in square only. [x. 73] 

Let it be contrived that the ratio of BE to DF is the same 
as the ratio of AB to CD ; [vi. 12} 

therefore also, as one is to one, so are all to all ; [v. t'a] 

therefore also, as the whole AE is to the whole CF, so is AB 
to CD. 

But AB is commensurable in length with CD. 
Therefore AE is also commensurable with CF, and BE 
with DF. [x. 11] 

And AE, EB are rational straight lines commensurable in 
square only ; 

therefore CF, FD are aJso rational straight lines commensur- 
able in square only. [x. 13] 

Now since, as AE is to CF, so is BE to DF, 
alternately therefore, as AE is to EB, so is CF to FD. [v. 16] 

And the square on AE is greater than the square on EB 
either by the square on a straight line commensurable with 
AE or by the square on a straight line incommensurable 
with it. 

If then the square on AE is greater than the square on 
EB by the square on a straight line commensurable with AE, 
the square on CF will also be greater than the square on FD 
by the square on a straight line commensurable with CF. 

[x. 14] 

And, if AE is commensurable in length with the rational 
straight line set out, 

CF is so also, [x. 1 3] 

if BE, then DF also, [id.] 

and, if neither of the straight lines AE, EB, then neither of 
the straight lines CF, FD. [x. 13] 

But, if the square on AE is greater than the square on EB 
by the square on a straight line incommensurable with AE, 

the square on CF will also be greater than the square on FD 
by the square on a straight line incommensurable with CF. 

[x. 14] 



x. io3, 104] PROPOSITIONS 103, 104 *3i 

And, if AE is commensurable in length with the rational 

straight line set out, 
CF is so also, 

if BE, then DFaho, [x. 13] 

and, if neither of the straight lines AE, EB, then neither of 
the straight lines CF, FD, [x. 13] 

Therefore CD is an apotome and the same in order 
with AB. 

Q. e. d. 

This and the following propositions to 107 inclusive (like the correspond- 
ing theorems x. 66 to 70) are easy and require no elucidation. They are 
equivalent to saying that, if in any of the preceding irrational straight lines 

- p is substituted for p, the resulting irrational is of the same kind and order 

as that from which it is altered. 



Proposition 104. 

A straight line commensurable with an apotome of a 
medial straight line is an apotome of a medial straight line 
and the same in order. 

Let AB be an apotome of a medial straight line, 
and let CD be commensurable in 

length with AB ; * .? & 

I say that CD is also an apotome of a (J B_F 

medial straight line and the same in 
order with AB. 

For, since AB is an apotome of a medial straight line, let 
EB be the annex to it. 

Therefore AE, EB are medial straight lines commensur- 
able in square only. [x. 74, 75] 

Let it be contrived that, as AB is to CD, so is BE to DF ; 

[vt. 12] 

therefore AE is also commensurable with CF, and BE 
with DF. [v. i2, x. 11] 

But AE, EB are medial straight lines commensurable in 
square only ; 

therefore CF, FD are also medial straight lines [x. 23] com- 
mensurable in square only ; [x. 13] 
therefore CD is an apotome of a medial straight line. [x. 74, 75] 






23a BOOK X [x. 104, 105 

I say next that it is also the same in order with AB. 
Since, as AE is to EB, so is CF to FD, 

therefore also, as the square on AE is to the rectangle AE, 
EB, so is the square on CF to the rectangle CF, FD. 

But the square on AE is commensurable with the square 
on CF; 

therefore the rectangle AE, EB is also commensurable with 
the rectangle CF, FD. [v. 16, x. 11] 

Therefore, if the rectangle AE, EB is rational, the rect- 
angle CF, FD will also be rational, [x. Dtf. 4' 

and if the rectangle AE, EB is medial, the rectangle CF, FD 

is also medial. [x. 23, Por.] 

Therefore CD is an apotome of a medial straight line and 

the same in order with AB. [x. 74, 75] 

Q. E. D. 

Proposition 105. 

A straight line commensurable with a minor straight line 
is minor. 

Let AB be a minor straight line, and CD commensurable 
with AB ; 
I say that CD is also minor. 

i A B E 

Let the same construction be made 

as before ; ? 2 F 

then, since AE, EB are incommensur- 
able in square, [x. 76] 
therefore CF, FD are also incommensurable in square, [x. 13] 

Now since, as AE is to EB, so is CF to FD, [v. 12, v. 16] 

therefore also, as the square on AE is to the square on EB, 
so is the square on CF to the square on FD. [vi. 22] 

Therefore, cumponendo, as the squares on AE, EB are to 
the square on EB, so are the squares on CF, FD to the 
square on FD. [v. 18] 

But the square on BE is commensurable with the square 
on DF; 

therefore the sum of the squares on AE, EB is also commen- 
surable with the sum of the squares on CF, FD. [v. 16, x. n] 

But the sum of the squares on AE, EB is rational ; [x. 76] 

therefore the sum of the squares on CF t FD is also rational. 

[x. Def. 4] 



x. io5, io6] PROPOSITIONS 104—106 133 

Again, since, as the square onAE is to the rectangle AE, 
EB, so is the squaru on CF to the rectangle CF, FD, 

while the square on AE is commensurable with the square 
on CF, 

therefore the rectangle AE, EB is also commensurable with 
the rectangle CF, FD. 

But the rectangle AE, EB is medial ; [x. 76] 

therefore the rectangle CF, FD is also medial ; [x. 13, Por.] 

therefore CF, FD are straight lines incommensurable in square 
which make the sum of the squares on them rational, but the 
rectangle contained by them medial. 

Therefore CD is minor. [x. j6] 

Q. E. D. 

Proposition 106, 

A straight tine commensurable with that "which produces 
with a rational area a medial whole is a straight line which 
produces with a rational area a medial whole. 

Let AB be a straight line which produces with a rational 
area a medial whole, 

and CD commensurable with AB ; . e E 
I say that CD is also a straight line 

which produces with a rational area a ' — ™ 

medial whole. 

For let BE be the annex to AB ; 

therefore AE, EB are straight lines incommensurable in 
square which make the sum of the squares on AE, EB 
medial, but the rectangle contained by them rational. [x. 77] 

Let the same construction be made. 

Then we can prove, in manner similar to the foregoing, 
that CF, FD are in the same ratio as AE, EB, 

the sum of the squares on AE, EB is commensurable with 
the sum of the squares on CF, FD, 

and the rectangle AE, EB with the rectangle CF, FD ; 

so that CF, FD are also straight lines incommensurable in 
square which make the sum of the squares on CF, FD medial, 
but the rectangle contained by them rational. 



»34 BOOK X [x. 106, 107 

Therefore CD is a straight line which produces with a 
rational area a medial whole. [x. 77] 

Q. E. D. 



Proposition ro7. 

A straight line commensurable with that which produces 
with a medial area a medial whole is itself also a straight line 
which produces with a medial area a medial whole. 

Let AB be a straight line which produces with a medial 
area a medial whole, 

and let CD be commensurable with AB; 

I say that CD is also a straight line " ' 

which produces with a medial area a ' 

medial whole. 

For let BE be the annex to AB, 

and let the same construction be made ; 

therefore AE, EB are straight lines incommensurable in 
square which make the sum of the squares on them medial, 
the rectangle contained by them medial, and further the sum 
of the squares on them incommensurable with the rectangle 
contained by them. [x. 78] 

Now, as was proved, AE, EB are commensurable with 
CF, ED, 

the sum of the squares on AE, EB with the sum of the 
squares on CF, FD, 

and the rectangle AE, EB with the rectangle CF, FD ; 

therefore CF, FD are also straight lines incommensurable in 
square which make the sum of the squares on them medial, 
the rectangle contained by them medial, and further the sum 
of the squares on them incommensurable with the rectangle 
contained by them. 

Therefore CD is a straight line which produces with a 
medial area a medial whole. [x. 78] 



x. io8] 



PROPOSITIONS 106—108 



*35 



Proposition 108. 








A e 




9 










c 

L 


G 






H 




> 


; 


F 



If from a rational area a medial area be subtracted, the 
"side" of the remaining area becomes one of two irrational 
straight lines, either an apotome or a minor straight line. 

For from the rational area BC let the medial area BD be 
subtracted ; 

I say that the " side " of the 
remainder EC becomes one 
of two irrational straight lines, 
either an apotome or a minor 
straight line. 

For let a rational straight 
line FG be set out, 
to FG let there be applied the 
rectangular parallelogram GH 
equal to BC, 

and let GK equal to DB be subtracted ; 
therefore the remainder EC is equal to LH. 

Since then BC is rational, and BD medial, 
while BC is equal to GH, and BD to GK, 
therefore GH is rational, and GK medial. 

And they are applied to the rational straight line FG ; 
therefore FH is rational and commensurable in length with 
FG, [x. 20] 

while FK is rational and incommensurable in length with FG; 

[X. 2 3 ] 

therefore FH is incommensurable in length with FK, [x. 13] 

Therefore FH, FK are rational straight lines commen- 
surable in square only ; 
therefore KH is an apotome [x. 73], and KF the annex to it. 

Now the square on HF is greater than the square on FK 
by the square on a straight line either commensurable with 
HF or not commensurable. 

First, let the square on it be greater by the square on a 
straight line commensurable with it. 

Now the whole HF is commensurable in length with the 
rational straight line FG set out ; 
therefore KH is a first apotome. [x. Deff. in. 1] 



i$6 BOOK X [x. 1 08, 109 

But the "side" of the rectangle contained by a rational 
straight line and a first apotome is an apotome. [x. 91] 

Therefore the " side " oiLH, that is, of EC, is an apotome. 

But, if the square on HF is greater than the square on 
FK by the square on a straight line incommensurable 
with HF, 

while the whole FH is commensurable in length with the 
rational straight line FG set out, 

KH is a fourth apotome. [x. Deff. in. 4] 

But the "side" of the rectangle contained by a rational 

straight line and a fourth apotome is minor. [x. 94] 

Q. E. D. 

A rational area being of the form kp*, and a medial area of the form 
Jk.p't the problem is to classify 

•J&p 1 - ,J\ . p* 
according to the different possible relations between k, K 
Suppose that <ru = Ap 1 , 

<rv = jk . p'. 
Since tru is rational and of medial, 
u is rational and « a-, 
while v is rational and v tr. 

Therefore u « v ; 

thus u, v are rational and au 
whence (u — v) is an apotome. 

The possibilities are now as follows. 
(1) «y«*-#* « u, 
(a) Vw* — 1? -j u. 
In both cases u « o-, 
so that (« — v) is either (1) a. first apotome, 
or (2) a fourth apotome. 
In case (r) •Joift — v) is an apotome [x. 91], 
but in case {2) V<r (u - v) is a m/ztur irrational straight line [x. 94]. 

Proposition 109. 

If from a medial area a rational area be subtracted, there 
arise two other irrational straight lines, either a first apotome 
of a medial straight line or a straight line which produces with 
a rational area a medial whole. 

For from the medial area BC let the rational area BD be 
subtracted. 






09] 



PROPOSITIONS 108, 109 



*37 



I say that the "side" of the remainder EC becomes one 
of two irrational straight lines, either a first apotome of a 
medial straight line or a straight line which produces with a 
rational area a medial whole. 



A 6 



F K H 

Q L 



For let a rational straight line FG be set out, 
and let the areas be similarly applied, 

It follows then that FH is rational and incommensurable 
in length with FG, 

while KF is rational and commensurable in length with FG ; 
therefore FH, FK are rational straight lines commensurable 
in square only ; [x. 13] 

therefore KH is an apotome, and FK the annex to it. [x. 73] 

Now the square on HF is greater than the square on FK 
either by the square on a straight line commensurable with 
HF or by the square on a straight line incommensurable 
with it. 

H then the square on HF is greater than the square on 
FK by the square on a straight line commensurable with HF, 
while the annex FK is commensurable in length with the 
rational straight line FG set out, 
KH is a second apotome. [x. Deff. in. a] 

But FG is rational ; 
so that the " side " of LH, that is, of EC, is a first apotome of 
a medial straight line. [x. 9a] 

But, if the square on HF is greater than the square on 
FK by the square on a straight line incommensurable with HF, 
while the annex FK is commensurable in length with the 
rational straight line FG set out, 

KH is a fifth apotome ; [x. Deff. in. 5] 

so that the "side" of EC is a straight line which produces 
with a rational area a medial whole. [x. 95] 



*3» 



BOOK X 



In this case we have to classify 
Suppose that av- Jk.ft, 



[x. ro9, no 



Thus, au being medial and av rational, 
u is rational and <j <r, 
while v is rational and « tr. 

Thus, as before, a, # are rational and «-, 
so that (w — v) is an apotome. 
Now either 

(r) Vw* — ^ * a, 
or (2) i/u'-tP ^ a, 

while in both cases » is commensurable with <r. 
Therefore (a — v) is either (1) a second apotome, 
or (z) a fifth apotome, 
and hence in case (1) vcr(« - v) is the. first apotome of a medial straight line, 

[*• 9*] 
and in case (a) \/<r' (a — v) is the "side" of a medial, minus a rational, area. 

[X-95] 

Proposition no. 

If from a medial area there be subtracted a medial area 
incommensurable with the whole, the two remaining irrational 
straight lines arise, either a second apotome of a medial straight 
line or a straight line which produces with a medial area a 
medial whole. 

For, as in the foregoing figures, let there be subtracted 
from the medial area BC the medial area BD incommensur- 
able with the whole ; 

F K H 



C 



I say that the " side " of EC is one of two irrational straight 
lines, either a second apotome of a medial straight line or a 
straight line which produces with a medial area a medial whole. 



x. no] PROPOSITIONS roo, no 239 

For, since each of the rectangles BC, BD is medial, 
and BC is incommensurable with BD, 

it follows that each of the straight lines FH, FK will be 
rational and incommensurable in length with FG. [x. m] 

And, since BC is incommensurable with BD, 
that is, GH with GK, 
HFh also incommensurable with FK; [vi. 1, x. n] 

therefore FH, FK are rational straight lines commensurable 
in square only ; 

therefore KH is an apotome. |x- 73] 

If then the square on FH is greater than the square on 
FK by the square on a straight line commensurable with FH, 

while neither of the straight lines FH, FK is commensurable 
in length with the rational straight line FG set out, 

KH is a third apotome. [x. Deff. m. 3] 

But KL is rational, 

and the rectangle contained by a rational straight line and a 
third apotome is irrational, 

and the "side" of it is irrational, and is called a second 
apotome of a medial straight line ; [x. 93] 

so that the "side " of LH, that is, of EC, is a second apotome 
of a medial straight line. 

But, if the square on FH is greater than the square on 
FK by the square on a straight line incommensurable with FH, 

while neither of the straight lines HF, FK is commensurable 
in length with FG, 

KH is a sixth apotome. [x. Deff. iti. 6] 

But the " side " of the rectangle contained by a rational 
straight line and a sixth apotome is a straight line which 
produces with a medial area a medial whole. [x. 96] 

Therefore the " side " of LH, that is, of EC, is a straight 
line which produces with a medial area a medial whole. 

Q. E. D. 

We have to classify <Jji . p* — ^/A . p\ 

where Jk . p" is incommensurable with J\ , p*. 

Put <TV=Jk. p>, 

<r» = JK . p>. 



2*0 



BOOK X 



[x. UO, III 



Then u is rational and v <r, 
v is rational and v <r, 
and u yd v. 

Therefore w, v are rational and «-. 
so that {u - v) is an apotome. 

Now either 

(l) 'Jtt t -V l « u, 
or {2) */w a - w 5 u u, 
while in both cases both u and z> are v o-. 
In case (1) {u~v) is a third apotome _ 
and in case (2) (« - ti) is a j/;e/^ apotome, 

sc that i/ir(B-0 is cither (1) a second apotome 0/ a medial straight Hne [x. 93], 
or (2) a "side" of the difference between two medial areas [x. 96]! 

Proposition hi. 

The apotome is not the same with the binomial straight line. 

Let AB be an apotome ; 
I say that AB is not the same with the 
binomial straight line 

For, if possible, let it be so ; 
let a rational straight line DC be set out, 
and tc CD let there be applied the 
rectangle CE equal to the square on 
AB and producing DE as breadth. 

Then, since AB is an apotome, 
DE is a first apotome. [x. 97] 

Let EF be the annex to it ; 
therefore DF, FE are rational straight 
lines commensurable in square only; 

the square on DF is greater than the square on FE by the 
square on a straight line commensurable with DF, 
and DF is commensurable in length with the rational straight 
line DC set out. [x. Deff. in. 1] 

Again, since AB is binomial, 
therefore DE is a first binomial straight line. [x. 60] 

Let it be divided into its terms at G, 
and let DG be the greater term ; 

therefore DG, GE are rational straight lines commensurable 
in square only, 




x. m] PROPOSITIONS no, m *4i 

the square on DG is greater than the square on GE by the 
square on a straight line commensurable with DG, and the 
greater term DG is commensurable in length with the rational 
straight line DC set out. - [s. Deff. h. i] 

Therefore DF is also commensurable in length with DG ; 

[x. ..] 

therefore the remainder GF is also commensurable in length 
with DF. [x. 15] 

But DF is incommensurable in length with EF\ 
therefore FG is also incommensurable in length with EF. [x. 13] 

Therefore GF, FE are rational straight lines commensur- 
able in square only ; 
therefore EG is an apotome. [x. 73] 

But it is also rational : 
which is impossible. 

Therefore the apotome is not the same with the binomial 
straight line. 

Q. E. D. 

This proposition proves the equivalent of the fact that 
Jx + Jy cannot be equal to *Jx - Jy', and 
x + Jy cannot be equal to x' - jy. 

We should prove these results by squaring the respective expressions; and 
Euclid's procedure corresponds to this exactly. 

He has to prove that 

p + ^i.p cannot be equal to p'—J\.p'. 

For, if possible, let this be so. 

Take the straight lines W*-?? , <?!: J±llZ . 
these must be equal, and therefore 

£(l+k+3j/b) = £(t + A- a Vfc) (i)- 

■ 1% 

Now — (1 + k\ — {1 + X) are rational and rt ; 



therefore [ p - (1 + A) - t (, + k)\ ~ £ (1 + A) 






w t tl :^, 



And, since both sides are rational, it follows that 

-(i+X)--(i+i)|--.2^ is an apotome. 



342 BOOK X [x. in 

But, by (i), this expression is equal to — . z Jk, which is rational. 

■ • 

Hence an apotome, which is irrational, is also rational: 
which is impossible. 

This proposition is the connecting link which enables Euclid to prove that 
all the compound irrationals with positive signs above discussed are different 
from all the corresponding compound irrationals with negative signs, while the 
two sets are all different from one another and from the medial straight line. 
The recapitulation following makes this clear. 



The apotome and the irrational straight lines following it 
are neither the same with the medial straight line nor with one 
another. 

For the square on a medial straight line, if applied to a 
rational straight line, produces as breadth a straight line 
rational and incommensurable in length with that to which it 
is applied, [x. 33] 

while the square on an apotome, if applied to a rational 
straight line, produces as breadth a first apotome, [x. 97] 

the square on a first apotome of a medial straight line, if 
applied to a rational straight line, produces as breadth a 
second apotome, [x. 98] 

the square on a second apotome of a medial straight line, if 
applied to a rational straight line, produces as breadth a third 
apotome, [x. 99] 

the square on a minor straight line, if applied to a rational 
straight line, produces as breadth a fourth apotome, [x. 100] 
the square on the straight line which produces with a rational 
area a medial whole, if applied to a rational straight line, 
produces as breadth a fifth apotome, [x. 101] 

and the square on the straight line which produces with a 
medial area a medial whole, if applied to a rational straight 
line, produces as breadth a sixth apotome. [x. 101] 

Since then the said breadths differ from the first and from 
one another, from the first because it is rational, and from one 
another since they are not the same in order, 
it is clear that the irrational straight lines themselves also 
differ from one another. 

And, since the apotome has been proved not to be the 
same as the binomial straight line, - [x. 1 j 1] 

but, if applied to a rational straight line, the straight lines 



x. in, ii a] PROPOSITIONS in, na *43 

following the apotome produce, as breadtKs, each according 
to its own order, apotomes, and those following the binomial 
straight line themselves also, according to their order, produce 
the binomials as breadths, 

therefore those following the apotome are different, and those 
following the binomial straight line are different, so that there 
are, in order, thirteen irrational straight lines in all, 

Medial, 

Binomial, 

First bimedial, 

Second bimedial, 

Major, 

"Side" of a rational plus a medial area, 

" Side " of the sum of two medial areas, 

Apotome, 

First apotome of a medial straight line, 

Second apotome of a medial straight line, 

Minor, 

Producing with a rational area a medial whole, 

Producing with a medial area a medial whole. 

Proposition 112. 

The square on a rational straight lint applied to the 
binomial straight line produces as breadth an apotome the 
terms of which are commensurable with the terms of the bi- 
nomial and moreover in the same ratio ; and further the 
apotome so arising will have the same order as the binomial 
straight line. 

Let A be a rational straight line, 
let BC be a binomial, and let DC be its greater term ; 
let the rectangle BC, EF be equal to the square on A ; 



I say that EFis an apotome the terms of which are commen- 
surable with CD, DB, and in the same ratio, and further EF 
will have the same order as BC. 



244 BOOK X [x. 112 

For again let the rectangle BD, G be equal to the square 
on A. 

Since then the rectangle BC, EF is equal to the rectangle 
BD, G, 
therefore, as CB is to BD, so is G to EF. [«. r6] 

But CB is greater than BD ; 
therefore G is also greater than EF. [v. 16, v. 14] 

Let EH be equal to G ; 

therefore, as CB is to BD, so is HE to EF; 

therefore, separando, as CD is to BD, so is HF to FE. [v. 17] 

Let it be contrived that, as HF is to FE, so is FK 
to KE; 

therefore also the whole HK is to the whole KF as FK 
is to KE ; 

for, as one of the antecedents is to one of the consequents, so 
are all the antecedents to all the consequents. [v. 12] 

But, as FK is to KE, so is CD to DB ; [v. 1 1] 

therefore also, as HK is to KF, so is CD to DB. [id.] 

But the square on CD is commensurable with the square 
on DB; [x, 36] 

therefore the square on HK is also commensurable with the 
square on KF. [vi. aa, x. 11] 

And, as the square on HK is to the square on KF, so is 
HK to KE, since the three straight lines HK, KF, KE are 
proportional. {v. Def. 9] 

Therefore HK is commensurable in length with KE, 

so that HE is also commensurable in length with EK. [x. 15] 

Now, since the square on A is equal to the rectangle 
EH,BD, 

while the square on A is rational, 

therefore the rectangle EH, BD is also rational. 

And it is applied to the rational straight line BD ; 

therefore EH is rational and commensurable in length 
with BD ; [x. 20] 

so that EK, being commensurable with it, is also rational and 
commensurable in length with BD. 



x. ira] PROPOSITION 112 345 

Since, then, as CD is to DB, so is FK to KE, 

while CD, DB are straight lines commensurable in square 
only, 

therefore FK, KE are also commensurable in square only, 

[x. 11] 

But KE is rational ; 

therefore FK is also rational. 

Therefore FK, KE are rational straight lines commen- 
surable in square only; 

therefore EF is an apotome. [x. 73] 

Now the square on CD is greater than the square on DB 
either by the square on a straight line commensurable with 
CD or by the square on a straight line incommensurable 
with it. 

If then the square on CD is greater than the square on 
DB by the square on a straight line commensurable with CD, 
the square on FK is also greater than the square on KE by 
the square on a straight line commensurable with FK. [x. 14] 

And, if CD is commensurable in length with the rational 
straight line set out, 

so also is FK; [x. u, 13] 

if BD is so commensurable, 

so also is KE ; [x. ia] 

but, if neither of the straight lines CD, DB is so commensur- 
able, 
neither of the straight lines FK, KE is so. 

But, if the square on CD is greater than the square on 
DB by the square on a straight line incommensurable 
with CD, 

the square on FK is also greater than the square on KE by 
the square on a straight line incommensurable with FK. [x. 14] 

And, if CD is commensurable with the rational straight 
line set out, 
so also is FK; 
if BD is so commensurable, 
so also is KE ; 



*46 BOOK X [x. 11* 

but, if neither of the straight lines CD, DJB is so commensur- 
able, 

neither of the straight lines FK t KE is so ; 
so that FE is an apotome, the terms of which FK, KE are 
commensurable with the terms CD, DB of the binomial 
straight line and in the same ratio, and it has the same order 
as^C 

Q. E. D. 

Heiberg considers that this proposition and the succeeding ones are inter- 
polated, though the interpolation must have taken place before Theon's time. 
His argument is that x. ti> — -115 are nowhere used, but that x. m rounds 
off the complete discussion of the 1 3 irrationals (as indicated in the recapitu- 
lation), thereby giving what was necessary for use in connexion with the 
investigation of the five regular solids. For besides x. 73 (used in xm. 6, n) 
x. 94 and 97 are used in xm. n, 6 respectively; and Euclid could not have 
stopped at x. 97 without leaving the discussion of irrationals imperfect, for 
X. 98 — ios are closely connected with x. 97,and x. 103 — 1 1 1 add, as it were, 
the coping-stone to the whole doctrine. On the other hand, X. n a— 115 are 
not connected with the rest of the treatise on the 13 irrationals and are not 
used in the stereometric books. They are rather the germ of a new study and 
a more abstruse investigation of irrationals in thansehts. Prop. 115 in 
particular extends the number of the different kinds of irrationals. As 
however x. n> — 115 are old and serviceable theorems, Heiberg thinks that, 
though Euclid did not give them, they may have been taken from Apollonius. 

I will only point out what seems to me open to doubt in the above, namely 
that x. us — 1 14 (excluding 115) are not connected with the rest of the 
exposition of the 13 irrationals. It seems tc me that they are so connected. 
x. 1 1 1 has shown us that a binomial straight line cannot also be an apotome. 
But X. ii2 — 1 14 show us how either of them "can be used to rationalise the other, 
thus giving what is surely an important relation between them. 

x. nz is the equivalent of rationalising the denominators of the fractions 
<* e* 

JA+JB' a+JB' 
by multiplying numerator and denominator by J A - JB and a — ,JB 
respectively. 

Euclid proves that — ,, — = \p - ^k . \p (h < 1 ), and his method enables 

us to see that \ = <r*/(p* - Ap 1 ). 

The proof is a remarkable instance of the dexterity of the Greeks in using 
geometry as the equivalent of our algebra. Like so many proofs in Archimedes 
and Apollonius, it leaves us completely in the dark as to how it was evolved. 
That the Greeks must have had some analytical method which suggested -the 
steps of such proofs seems certain ; but what it was must remain apparently 
an insoluble mystery. 

I will reproduce by means of algebraical symbols the exact course of 
Euclid's proof. 

He has to prove that fi — is an apotome related in a certain way to 

p + J* . p 



x. m] PROPOSITION lit S47 

the binomial straight litiu p + Jk . p. If « be the straight line required, 
(u + w) - w is shown to be an apotome of the kind described, where w is 
determined in the following manner. 

We have (p + Ji.p) u = tr*= Ji .p. x, say, 1 
whence x > u. \ (i). 

Let x = a + v. I 

Then (p + Ji . p) : Jk . p = (» + *>): a, 

and hence p : Jk.p = v : a (2). 

■ Let w be taken such that 

v \ u = (a + w) : to (3). 

Thus v : a = (a + y + to) \ (k + to) (4), 

and therefore p : Ji . p = (a + v + to) : (a + to). 

From the last proportion, 

(a + v + to)* « (a + to)*, 
and, from the two preceding, (« + to) is a mean proportional between 
(u + v +■ to), to, so that 

{a + v + to)* : (a + to)* = (a + » +■ to) : w. 

Therefore (a +■ y + to) « to, 

whence (a + 1*) « «/. 

Now Jk . p (a + v) = a 1 , which is rational ; 
therefore (a + v) is rational and " Jk . p; 
hence to is also rational and rt Jk-p • (S)- 

Next, by (a), (3), since p, Jk . p are «- , 
(a + w) «- to, 
and to is rational , 

therefore (a + to) is rational, 

and (« + to), to are rational and is* . 

Hence (a + to) - to is an apotome. 

Now either (I) Jp*-Ap' r> p, 

or (II) 4f?-*# - p. 

In case (I) V(a + iv)*-v? « {« + a>), [(z), (3) and X. 14] 

and in case (II) V(a + to)* - to 1 ^ (a + to). [«£] 

Then, since [(5)] w n Jk.p, 
by X. 1 1 and (2), (3), (a + to) - p (6). 

[This step is omitted in Euclid, but the result is assumed.] 

If therefore p « cr, (a +to) " or; 
if JA.p « cr, hi «<r; [(5)] 

and, if neither p nor Ji . p is n <r, neither (a + to) nor to will be " <r. 

Thus the order of the apotome (a + to) - w is the same as th^t of the 
binomial straight line p+ Jk.p; while [(2), (3)] the terms are proportional 
and [(5), (6)] commensurable respectively. 



j 4 8 BOOK X 

We find (u + w),w algebraically thus. 
u 



By (i), 

and. by (a), (3), 
whence 

Thus 

Therefore 



p * Jk . p ' 
« + w _ p 

U.Jk.p 

p-Jk.p 



U + «' = If 



If'. 


^- 


c 




P 5 


-V 






I 


<r> 


■ P 


re . 


7*~ 


P*- 


"V 



(« + «:)- W = a" . — 3 



p-Jk.p 



[X. 112, 113 



p'-V 



Proposition 113. 

7^ square on a rational straight line, if applied to an 
apotome, produces us- breadth the binomial straight line the 
terms of which are commensurable with the terms of the 
apotome and in the same ratio ; and further the binomial 
so arising has the same order as the apotome. 

Let A be a rational straight line and BD an apotome, 
and let the rectangle BD, KH be equal to 
the square on A, so that the square on the 
rational straight line A when applied to the 
apotome BD produces KH as breadth ; 
I say that KH is a binomial straight line the 
terms of which are commensurable with the 
terms of BD and in the same ratio ; and 
further KH has the same order as BD. 

For let DC be the annex to BD ; 

therefore BC, CD are rational straight lines commensurable 
in square only. [x. 73] 

Let the rectangle BC, G be also equal to the square on A . 
But the square on A is rational ; 

therefore the rectangle BC, G is also rational. 

And it has been applied to the rational straight line BC ; 

therefore G is rational and commensurable in length with BC 

[x. so] 



x. its] PROPOSITIONS 112, 113 249 

Since now the rectangle BC, G is equal to the rectangle 
BD, KH, 

therefore, proportionally, as CB is to BD, so is KH to G. 

[VI. 16] 

But BC is greater than BD ; 
therefore KH is also greater than G. [v. 16, v. 14] 

Let KE be made equal to G ; 
therefore KE is commensurable in length with BC. 

And since, as CB is to BD, so is HK to KE, 
therefore, convertendo, as BC is to CD, so is .A'/f to HE. 

[v. 19, Por.] 

Let it be contrived that, as KH is to HE, so is HE 
to EE; 

therefore also the remainder KE is to EH as KH is to HE, 
that is, as .ffC is to CD. [v. 19] 

But BC, CD are commensurable in square only ; 

therefore KE, EH are also commensurable in square only. 

[x. 11] 
And since, as KH is to HE, so is KE to EH, 
while, as KH is to HE, so is HE to /vfi", 
therefore also, as KE is to ^.#, so is HE to /^-fi", [v. it] 

so that also, as the first is to the third, so is the square on the 
first to the square on the second ; [v. Def. 9] 

therefore also, as KE is to EE, so is the square on KE to the 
square on EH. 

But the square on KE is commensurable with the square 
on EH, 

for KE, EH are commensurable in square ; 

therefore KE is also commensurable in length with EE, [x. n] 

so that KE is also commensurable in length with KE. [x. 15] 

But KE is rational and commensurable in length with BC; 

therefore KE is also rational and commensurable in length 
with^C. [X. !2] 

And, since, as BC Is to CD, so is KE to EH, 
alternately, as BC is to KE, so is DC to EH. [v. 16] 

But BC is commensurable with KE; 
therefore EH is also commensurable in length with CD. fx. n] 



aS<> BOOK X [x. 113 

But BC, CD are rational straight lines commensurable in 
square only ; 

therefore KF, FH are also rational straight lines [x. Def. 3] 

commensurable in square only ; 

therefore KH is binomial. [x. 36] 

If now the square on BC is greater than the square on CD 
by the square on a straight line commensurable with BC, 
the square on KF will also be greater than the square on FH 
by the square on a straight line commensurable with KF, [x 14] 

And, if BC is commensurable in length with the rational 
straight line set out, 
so also is KF ; 

if CD is commensurable in length with the rational straight 
line set out, 
so also is FH, 

but, if neither of the straight lines BC, CD, 
then neither of the straight lines KF, FH. 

But, if the square on BC is greater than the square on CD 
by the square on a straight line incommensurable with BC, 
the square on KF is also greater than the square on FH by 
the square on a straight line incommensurable with KF. [x. 14] 

And, if BC is commensurable with the rational straight 
line set out, 
so also is KF ; 
if CD is so commensurable, 
so also is FH ; 

but, if neither of the straight lines BC, CD, 
then neither of the straight lines KF, FH. 

Therefore KH is a binomial straight line, die terms of 
which KF, FH are commensurable with the terms BC, CD of 
the apotome and in the same ratio, 
and further KH has the same order as BD. 

Q. E. D. 

This proposition, which is companion to the preceding, gives us the equiva- 
lent of the rationalisation of the denominator of 

JA-JB 0T a-JB' 



x. n 3 ] PROPOSITION 113 151 

Euclid {or the writer) proves that 
_i 
Ji —= i Xf>*XJA.p, (t<t) 

and his method enables us to see that X = <r*/(p* - kp 1 ). 

Let 7T~ n = * '• 

p-Jk.p 

and it is proved that u is the binomial straight line (» - «') + w, where w is 

determined as shown below. 

u {p -. Jk . p) = <r* = px, say, 

whence p : (p — JA ,p) = u • x (1), 

so that x •■-. u. 

Let then x = a - v. 

Since (u-v)p = a 1 , a rational area, 

(u-v) is rational and " p (*). 

And [(1)] p:(p-JA.p) = u: (w-v), 

so that, cenverfendoy p : Jk . p = u : v. 

Suppose that « : v = w : (v - w), 

so that [v. 19] {« - w) : w = u : v = tu : (v - w). 

Thus, w being a mean proportional between (u - w), (f - w), 
{» — wf : w 1 = (a — w) : {v - w). 

But {u — ntf : it?= li 1 v& 

-«**V (3). 

so that (u — 10)* *» w*. 

Therefore (« - (c) « (w - a>) 

" {{» - w) - (v - w)\ 
*(u- v). 

Therefore [(2)] (u-w) is rational and « p (4)- 

And, since p: Jk. p = (u-w):v>, 

w is rational and^^.p (5). 

Hence [(4}, (5)] (u-w), w are rational and «-, 
so that (k - w) + w is a binomial straight line. 

Now either (I) <Jp* - kp* « p, 



or (II) Vp* - V " P- 

Incase (I) \/{« - a/)* - w'* « (« — a>), 

and in case (II) V(« - w)* - «/" « (»-«>). [(3) and x. 14] 

And, if p " tr, (u - w) « o- ; [{4)] 

if ,/A . p " er, a/ " <r ; [5] 

while, if neither p nor ,/i. p is « 0-, neither {u-w) nor w is " a. 

Hence (u-w) + if is a binomial straight line of the same order as the 

apotome p — J A . />, its terms are proportional to those of the -"wtome [{3)], 

and commensurable with them respectively [{4), {5)]. 



*5* 



BOOK X 




[x. 113, 114 


To find (a - w), w algebraically we have 






<r* 






tt -p-Jk.p' 




u -if p 






w ~ Jk.p' 




From the latter w - —'-,.'- 

P + Jk . P 






a 1 . Jk ,p 

Thus u - w - w ■ -n = . 

Jk p- 






Therefore (« - »/) + w = v* . p -^^. \ P ■ 





p*-v 



Proposition 114. 

If an area be contained by an apotome and the binomial 
straight line the terms of which are commensurable with the 
terms of the apotome and in the same ratio, the " side " of the 
area is rational. 

For let an area, the rectangle AB, CD, be contained by 
the apotome AB and the binomial 
straight line CD, 

A B F 

and let CE be the greater term of 

the latter; C E_ p 

let the terms CE, ED of the o_ 

binomial straight Hne be commen- H 
surable with the terms AF, FB of 

the apotome and in the same ratio; * 1 *? 

and let the "side" of the rectangle 
AB, CBbeG; 

I say that G is rational. 

For let a rational straight line H be set out, 
and to CD let there be applied a rectangle equal to the square 
on //"and producing KL as breadth. 

Therefore KL is an apotome. 

Let its terms be KM, ML commensurable with the terms 
CE, ED of the binomial straight line and in the same ratio. 

[x. ,„] 



x. u 4 ] PROPOSITIONS 113,114 253 

But CE, ED are also commensurable with AF, FB and in 
the same ratio ; 

therefore, as AF is to FB, so is KM to ML. 

Therefore, alternately, as AF is to KM, so is BF to L M ; 
therefore also the remainder AB is to the remainder KL as 
AF is to KM. [v. 19] 

But AFis commensurable with KM; [x. ra] 

therefore AB is also commensurable with KL. [x. n] 

And, as AB is to KL, so is the rectangle CD, AB to the 
rectangle CD, KL ; [vi. 1] 

therefore the rectangle CD, AB is also commensurable with 
the rectangle CD, KL. [x. u] 

But the rectangle CD, KL is equal to the square on H ; 
therefore the rectangle CD, AB is commensurable with the 
square on H. 

But the square on G is equal to the rectangle CD, AB ; 
therefore the square on G is commensurable with the square 
on H. 

But the square on H is rational ; 

therefore the square on G is also rational ; 

therefore G is rational. 

And it is the "side" of the rectangle CD, AB. 
Therefore etc. 

Porism. And it is made manifest to us by this also that 
it is possible for a rational area to be contained by irrational 
straight lines. 

Q. E. D. 

This theorem is equivalent to the proof of the fact that 

J(JA - JB) OQA + A JB) = Jk{A-B), 

W»d JJa ~ JB) (ka + A JB) = Jk{^B). 

The result of the theorem x. 1 1 2 is used for the purpose thus. 

We have to prove that 

■J(j>-Jk.p){\p + \Jk. P ) 
is rational. 

By X. 11a we have, if a is a rational straight line, 



Xp + A Ji . p 



»--*>- *V*-P (0- 



a 54 BOOK X [x. 114, 115 

Now p : A'p = Jk . p : X' Jk . p = (p - Jk . p) : (A'p - A' Jk . p), 
so that (p - Jk . p) n {A'p - X' Jk . p). 

Multiplying each by (Xp + A Jk . p), we have 

(p-Jk.p)()* + \Jk.p)~(\p + XJk.p)(\'p-k , Jk.p) 
- <r*, by <r). 

That is, (p- Jk .p)(\p + \Jk.p) is a rational area, 
and therefore V(p - Jk. p) (Ap + kjk.p) is rational. 

Proposition 115. 

From a medial straight line there arise irrational straight 
lines infinite in number, and none of them is the same as any 
of the preceding. 

Let /4bea medial straight line ; 
I say that from A there arise 

irrational straight lines infinite in A , 

number, and none of them is the 
same as any of the preceding. 



Let a rational straight line B c 

be set out, 

and let the square on C be equal 

to the rectangle B, A ; 

therefore C is irrational ; [x. Def. 4] 

for that which is contained by an irrational and a rational 

straight line is irrational. [deduction from x. 20] 

And it is not the same with any of the preceding ; 
for the square on none of the preceding, if applied to a rational 
straight line produces as breadth a medial straight line. 

Again, let the square on D be equal to the rectangle B, C; 
therefore the square on D is irrational. [deduction from x. ao] 

Therefore D is irrational ; [x, Def. 4] 

and it is not the same with any of the preceding, for the 
square on none of the preceding, if applied to a rational 
straight line, produces C as breadth. 

Similarly, if this arrangement proceeds ad infinitum, it 
is manifest that from the medial straight line there arise 
irrational straight lines infinite in number, and none is the 
same with any of the preceding. 

Q. E. D. 



x. iij] PROPOSITIONS 114, 115 «55 

Heiberg is clearly right in holding that this proposition, at all events, is 
alien to the general scope of Book x, and is therefore probably an interpola- 
tion, made however before Theon's time. It is of the same character as a 
scholium at the end of the Book, which is (along with the interpolated proposi- 
tion proving, in two ways, the incommensurability of the diagonal of a square 
with its side) relegated by August as well as Heiberg to an Appendix. 

The proposition amounts to this. 

The straight line k*p being medial, if o- be a rational straight line, v &pv 
is a new irrational straight line. So is the mean proportional between this 
and another rational straight line <r\ and so on indefinitely. 

Ancient Extensions of the Theory of Book X. 

From the hints given by the author of the commentary found in Arabic 
by Woepcke (cf. pp. 3 — 4 above) it would seem probable that Apollonius' 
extensions of the theory of irrationals took two directions : (1 ) generalising 
the medial straight line of Euclid, and (2) forming compound irrationals by the 
addition and subtraction of more than two terms of the sort composing the 
binomials, apotomes, etc The commentator writes (Woepcke's article, pp. 694 
sqq.): 

"It is also necessary that we should know that, not only when we join 
together two straight lines rational and commensurable in square do we obtain 
the binomial straight line, but three or four lines produce in an analogous 
manner the same thing. In the first case, we obtain the trinomial straight 
line, since the whole line is irrational ; and in the second case we obtain the 
quadrinomial, and so im si infinitum. The proof of the (irrationality of the) 
line composed of three lines rational and commensurable in square is exactly 
the same as the proof relating to the combination of two lines. 

" But we must start afresh and remark that not only can we take one sole 
medial line between two lines commensurable in square, but we can take three 
or four of them and so on ad infinitum, since we can take, between any two 
given straight lines, as many lines as we wish in continued proportion. 

" Likewise, in the lines formed by addition not only can we construct the 
binomial straight line, but we can also construct the trinomial, as well as the 
first and second tri medial ; and, further, the line composed of three straight 
lines incommensurable in square and such that the one of them gives with 
each of the two others a sum of squares (which is) rational, while the rectangle 
contained by the two lines is medial, so that there results a major (irrational) 
composed of three lines. 

"And, in an analogous manner, we obtain the straight line which is the 
' side ' of a rational plus a medial area, composed of three straight lines, and, 
likewise, that which is the ' side ' of (the sum of) two medials. " 

The generalisation of the medial is apparently after the following manner. 
Let x, y be two straight lines rational and commensurable in square only and 
suppose that m means are interposed, so that 

* : x, - Jt) : x, = x, : x$ = . . . = x m - t ; x m -x m : y. 



We easily derive herefrom — = I — I , 
x r \xj 

~ w 



x 

y 



*5« BOOK X 

and hence *,' = x T . x r J ', 

so that (jf r .* r - , ) l " + 1 = (>.jc" , )'' 1 

and therefore x,™* 1 = *" -F+1 .y, 

or AV = (*"- r + '>T+ r , 

which is the generalised medial. 

We now pass to the trinomial etc., with the commentator's further remarks 
about them. 

(i) Tht trinomial. "Suppose three rational straight lines commensurable in 
square only. The line composed of two of these lines, that is, the binomial 
straight line, is irrational, and, in consequence, the area contained by this line 
and the remaining line is irrational, and, likewise, the double of the area 
contained by these two lines will be irrational. Thus the square on the 
whole line composed of three lines is irrational and consequently the line is 
irrational, and it is called a trinomial straight line." 

It is easy to see that this "proof" is not conclusive as stated. Nor does 
Woepcke seem to show how the proposition can be proved on Euclidean 
lines. But I think it would be somewhat as follows. 

Suppose x, y, z to be rational and '*- . 

Then «*, y*, «* are rational, and zyz, zzx, zxy are all medial. 

First, (zyz + zzx + 2xy) cannot be rational. 

For suppose this sum equal to a rational area, say a 3 . 

Since zyz +■ zzx + zxy = ff 1 , 

zzx + zxy = <r* - zy z, 

or the sum of two medial areas incommensurable with one another is equal to 
the difference between a rational area and a medial area. 

But the " side" of the sum of the two medial areas must [x. 72] be one of 
two irrationals with a positive sign ; and the " side " of the difference between a 
rational area and a medial area must [x. 108] be one of two irrationals with a 
negative sign. 

And the first " side " cannot be the same as the second [x. 1 1 1 and ex- 
planation following]. 

Therefore zzx + zxy ♦ cr" — zyz, 

and zyz + zzx + zxy is consequently irrational. 

Therefore {x 1 +>" + «*) u (zyz + zex + zxy), 

whence (x+y + *)* ~ (*■ +y* +■ 3'), 

so that (x + y + zf, and therefore also (x +y + z), is irrational. 

The commentator goes on : 

" And, if we have four lines commensurable in square, as we have said, the 
procedure will be exactly the same ; and we shall treat the succeeding lines in 
an analogous manner." 

Without speculating further as to how the extension was made to the 
quadrinsmial etc., we may suppose with Woepcke that Apollonius probably 
investigated the multinomial 

p+ Jk. p + ,J\.p+ Jp.p + ... 



ANCIENT EXTENSIONS »S7 

(2) The first trimedial straight line. 

The commentator here says : "Suppose we have three medial lines com- 
mensurable in square [only], one of which contains with each of the two others 
a rational rectangle ; then the straight line composed of the two lines is 
irrational and is called the first bi medial ; the remaining line is medial, and 
the area contained by these two lines is irrational. Consequently the square 
on the whole line is irrational." 

To begin with, the conditions here given are incompatible. If x, y, z be 
medial straight lines such that xy, xz are both rational, 

y : z = xy : xz = m : n, 

and y, i are commensurable in length and not in square only. 

Hence it seems that we must, with Woepcke, understand " three medial 

straight lines such that one is commensurable with each of the other two in 

square only and makes with it a rational rectangle." 
If x, y, i be the three medial straight lines, 

so that {?? +y + s') is medial. 

Also we have txy, 2x1 both rational and 2ys medial. 

Now (x' + y 1 + sfj + tyt + ixy + txz cannot be rational, for, if it were, the 
sum of two medial areas, {x i +y t + «*), ays, would be rational: which is im- 
possible, [Cf, x. 7 a.] 

Hence (x +y + z) is irrational. 

(3) The second trimedial straight line. 

Suppose x, y, z to be medial straight lines commensurable in square only 
and containing with each other medial rectangles. 

Then (x 1 +_v* + «•) « x*, and is medial. 

Also tyz, 2zx, txy are all medial areas. 

To prove the irrationality in this case I presume that the metnod would 
be like that of x. 38 about the second bimedial. 
Suppose o- to be a rational straight line and let 

(x' + f + z') = <rt 
tyz = ITU 
22X = <rv 
3xy = aw ] 

Here, since, e.g., xz : xy - v : w, 

or ss\y = v:w, 

and similarly x :z -w :v, 

u, v, w are tommensurable in square only. 

Also, since {x 1 + >* + 1*) " x* 

w xy 
t is incommensurable with w. 



aj8 BOOK X 

Similarly / is incommensurable with u, v. 

But t, u, v, to are all rational and "- er. 

Therefore (t + u + v+w) is a quadrinomial and therefore irrational. 
Therefore <r (/ + « + v+ w), or (# +,y + *)*, is irrational, 
whence (x +y + *) is irrational. 

{4) The major made up of three straight lines. 

The commentator describes this as "the line composed of three straight 
lines incommensurable in square and. such that one of them gives with each 
of the other two a sum of squares (which is) rational, while the rectangle 
contained by the two lines is medial." 

If *, y, x are the three straight lines, this would indicate 

(jc'+y) rational, 

(jc 1 +■ **) rational, 

tyx medial. 

Woepcke points out (pp. 696— 8, note) the difficulties connected with this 
supposition or the supposition of 

(«*+^) rational, 
(s? + e 1 ) rational, 
ixy (or ixt) medial, 
and concludes that what is meant is the supposition 

(«* +jP) rational 
xy medial 
xz medial , 
(though the text is against this). 

The assumption of (j.- 1 +y') and (a^ + **) being concurrently rational is 
certainly further removed from Euclid, for x. 33 only enables us to find one 
pair of lines having the property, as x, y. 

But we will not pursue these speculations further- 
As regards further irrationals formed by subtraction the commentator 
writes as follows. 

" Again, it is not necessary that, in the irrational straight lines formed by 
means of subtraction, we should confine ourselves to making one subtraction 
only, so as to obtain the apotome, or the first apotome of the medial, or the 
second apotome of the medial, or the minor, or the straight line which 
produces with a rational area a medial whole, or that which produces with a 
medial area a medial whole ; but we shall be able here to make two or three 
or four subtractions. 

"When we do that, we show in manner analogous to the foregoing that 
the lines which remain are irrational and that each of them is one of the lines 
formed by subtraction. That is to say that, if from a rational line we cut off 
another rational line commensurable with the whole line in square, we obtain, 
for remainder, an apotome ; and, if we subtract from this line (which is) 
cut off and rational — that which Euclid calls the antux (irpowofj/uifowra) — 
another rational line which is commensurable with it in square, we obtain, as 
the remainder, an apotome ; likewise, if we cut off from the rational line cut 



ANCIENT EXTENSIONS 



*59 



off from this line (i.e. the annex of the apotome last arrived at) another line 
which is commensurable with it in square, the remainder is an apotome. The 
same thing occurs in the subtraction of the other lines." 

As Woepcke remarks, the idea is the formation of the successive apotomes 
,/a - Ji, jb — Jc, Je— Jd t etc. We should naturally have expected to see 
the writer form and discuss the following expressions 















BOOK XL 

DEFINITIONS, 
i. A solid is that which has length, breadth, and depth. 

2. An extremity of a solid is a surface. 

3. A straight line is at right angles to a plane, 
when it makes right angles with all the straight lines which 
meet it and are in the plane. 

4. A plane is at right angles to a plane when the 
straight lines drawn, in one of the planes, at right angles to 
the common section of the planes are at right angles to the 
remaining plane. 

5. The inclination of a straight line to a plane 
is, assuming a perpendicular drawn from the extremity of 
the straight line which is elevated above the plane to the 
plane, and a straight line joined from the point thus arising 
to the extremity of the straight line which is in the plane, 
the angle contained by the straight line so drawn and the 
straight line standing up. 

6. The inclination of a plane to a plane is the acute 
angle contained by the straight lines drawn at right angles 
to the common section at the same point, one in each of the 
planes. 

7. A plane is said to be similarly inclined to a plane 
as another is to another when the said angles of the inclina- 
tions are equal to one another, 

8. Parallel planes are those which do not meet. 



xi. Dkff. 9— 19] DEFINITIONS 1— 19 *6i 

9. Similar solid figures are those contained by similar 
planes equal in multitude. 

10. Equal and similar solid figures are those con- 
tained by similar planes equal in multitude and in magnitude, 

11. A solid angle is the inclination constituted by more 
than two lines which meet one another and are not in the 
same surface, towards all the lines. 

Otherwise : A solid angle is that which is contained by 
more than two plane angles which are not in the same plane 
and are constructed to one point. 

12. A pyramid is a solid figure, contained by planes, 
which is constructed from one plane to one point. 

13. A prism is a solid figure contained by planes two 
of which, namely those which are opposite, are equal, similar 
and parallel, while the rest are parallelograms. 

14. When, the diameter of a semicircle remaining fixed, 
the semicircle is carried round and restored again to the same 
position from which it began to be moved, the figure so 
comprehended is a sphere. 

15. The axis of the sphere is the straight line which 
remains fixed and about which the semicircle is turned. 

16. The centre of the sphere is the same as that 
of the semicircle, 

17. A diameter of the sphere is any straight line 
drawn through the centre and terminated in both directions 
by the surface of the sphere, 

18. When, one side of those about the right angle in a 
right-angled triangle remaining fixed, the triangle is carried 
round and restored again to the same position from which it 
began to be moved, the figure so comprehended is a cone. 

And, if the straight line which remains fixed be equal to 
the remaining side about the right angle which is carried 
round, the cone will be right-angled ; if less, obtuse-angled ; 
and if greater, acute-angled. 

19. The axis of the cone is the straight line which 
remains fixed and about which the triangle is turned. 



26a BOOK XI [xi. Dkff. 

20. And the base is the circle described by the straight 
line which is carried round. 

21. When, one side of those about the right angle in a 
rectangular parallelogram remaining fixed, the parallelogram 
is carried round and restored again to the same position from 
which it began to be moved, the figure so comprehended is a 
cylinder. 

22. The axis of the cylinder is the straight line which 
remains fixed and about which the parallelogram is turned. 

23. And the bases are the circles described by the two 
sides opposite to one another which are carried round. 

24. Similar cones and cylinders are those in which 
the axes and the diameters of the bases are proportional. 

25. A cube is a solid figure contained by six equal 
squares. 

26. An octahedron is a solid figure contained by eight 
equal and equilateral triangles. 

27. An icosahedron is a solid figure contained by 
twenty equal and equilateral triangles. 

28. A dodecahedron is a* solid figure contained by 
twelve equal, equilateral, and equiangular pentagons. 

Definition i. 

%rtp*6v tart to ftyKos *<u irAaTO? ko\ fjafios *X nv - 

This definition was evidently traditional, as may be inferred from a number 
of passages in Plato and Aristotle. Thus Plato speaks (Sophist, 235 d) of 
making an imitation of a model (irapdSiiy(ia) " in length and breadth and 
depth and (Lams, 8^7 e) of "the art of measuring length, surface and depth" 
as one of three fiadr^aro. Depth, the third dimension, is used alone as a 
description of " body " by Aristotle, the term being regarded as connoting the 
other two dimensions ; thus (Afttaph. 1020 a 13, 11) "length is a fine, breadth a 
surface, and depth body" ; " that which is continuous in one direction is length, 
in two directions breadth, and in three depth." Similarly Plato (Rep. 528 b, d), 
when reconsidering his classification of astronomy as next to (plane) geometry: 
"although the science dealing with the additional dimension of depth is next in 
order, yet, owing to the fact that it is studied absurdly, I passed it over and 
put next to geometry astronomy, the motion of (bodies having) depth." In 
Aristotle (Topia vi. 5, 142 b 24) we find "the definition of body, that which 
has three dimensions (itafrraatis:) " ; elsewhere he speaks of it as " that which 
has all the dimensions " (Dt catlo 1. 1, 268 b 6), "that which has dimension 
every way" (™ vivrrj Buurrao-ii' cyov, Afttaph. 1066 b 32) etc. In the Physics 



xi. Dbff. i— s] DEFINITIONS AND NOTES 263 

(iv. 1, 208 b i3sqq.) he speaks of the " dimensions " as six, dividing each of 
the three into two opposites, "up and down, before and behind, right and left," 
though of course, as he explains, these terms are relative. 

Heron, as might be expected, combines the two forms of the definition. 
41 A solid body is that which has length, breadth, and depth : or that which 
possesses the three dimensions." (Def. 1 1 .) 

Similarly Theon of Smyrna (p. 1 1 1, 19, ed. Hiller) : " that which is extended 
(Suurrarav) and divisible in three directions is solid, having length, breadth 
and depth." 

Definition 2. 

In like manner Aristotle says (Metaph. 1066 b 13) that the notion (Ady«) 
of body is "that which is bounded by surfaces" («rtiri£o« in this case) and 
{Metaph. 1060 b 15) "surfaces {brup&viuu) are divisions of bodies." 

So Heron (Def. 11.): " Every solid is bounded (Tripa.rovro,i) by surfaces, and 
is produced when a surface is moved from a forward position in a backward 
direction." 

Definition 3. 
1 J 

EiWtla irpot ttmwm* op6^ tarty, Srav irpos irdtras ras durofttras aur^s tiQttas 
KaX atkraf iv t£ iiti-riStf ApSas ttoijj ywt'a?. 

This definition and the next are given almost word for word by Heron 
(Def. 115). 

That a straight line can be so related to a plane as described in Def. 3 is 
established in XI. 4. The fact has been made the basis of a definition of a 
plane which is attributed by Crelle to Fourier, and is as follows. " A plane is 
formed by the totality of all the straight lines which, passing through one and 
the same point of a straight line in space, stand perpendicular to it," Stated 
in this form, the definition is open to the objection that the conception of a 
right angle, involving the measurement of angles, presupposes a plane, inasmuch 
as the measurement of angles depends ultimately upon the superposition of two 
planes and their coincidence throughout when two lines in one coincide with 
two lines in the other respectively. Cf. my note on 1. Def. 7, Vol. 1. pp. 1 73 — 5. 

Definition 4. 

'EtrinSor irpos tnirtSov ip$6v itrnv, oto.v at rg itotvj) To/t]j iw twari&uv -rpos 
6p8a$ ayofuvai ntfcuu iv ivl tcuv ^imrtoW T$ Xoifftp i-rtiire&tv wpot 6p$as ut&tv. 

Both this definition and Def. 6 use the common section of two planes, 
though it is not till xi. 3 that this common section is proved to be a straight 
line. The definition however, just like Def, 3, is legitimate, because the object 
is to explain the meaning of terms, not to prove anything 

The definition of perpendicular planes is made by Legendre a particular 
case of Def. 6, the limiting case, namely, where the angle representing the 
" inclination of a plane to a plane " is a right angle. 

Definition 5. 

EAoV&n rpot iriwtSov «Aor« ioriv, orav djro toC /Lcrrupav Wparos nj? 
tiOtlas iirl TO tvartBov t<a&tTO<; d)^jj, * al *^ 7 ™ T °^' ywoptvav O^Tjpttiov iwl to iv rid 
luilfffg Tripat rffc tiBttas tuStia iwi£tvx&g, 17 irfpif\0)Urjj ywvia viro r!js ixSturtft 

(a! Tljl (^KffTWOTJt. 



*64 BOOK XI [xi. Dwr. 5, 6 

In other words, the inclination of a straight line to a plane is the angle 
between the straight line and its projection on the plane. This angle is of 
course less than the angle between the straight line and any other straight line 
in the plane through the intersection of the straight line and plane ; and the 
fact is sometimes made the subject of a proposition in modern text-books. It 
is easily proved by means of the propositions xi. 4, 1. 19 and 18. 

Definition 6. 

'EffiWSou irpos tiTLTrtBov jcAuto itrrlv t} Trtpif^o^rtyy} o£tia. ywta vwb rwv irpos 
opOas Tfj Koivfj Top-?/ dyopfywv irpos ty aurtu trqprtiw Iv tKaripw twv iiniriZtov. 

When two planes meet in a straight line, they form what is called in 
modern text-books a dihedral angle, which is defined as the opening or angular 
opening between the two planes. This dihedral angle is an " angle " altogether 
different in kind from a plane angle, as again it is different from a solid angle 
as defined by Euclid (i.e. a trihedral, tetrahedral, etc. angle). Adopting for 
the moment Apollonius' conception of an angle as the "bringing together of a 
surface or solid towards one point under a broken line or surface " (Proclus, 
p. 123, 16}, we may regard a dihedral angle as the bringing together of the 
broken surface formed by two intersecting planes not to a point but to a straight 
line, namely the intersection of the planes. Legend re, in a proposition on the 
subject, applied provisionally the term corner to describe the dihedral angle 
between two planes ; and this would be a better word, I think, than opening 
to use in the definition. 

The distinct species of " angle " which we call dihedral is, however, 
Measured by a certain plane angle, namely that which Euclid describes in the 
present definition and calls the inclination 0/ a plane to a plane, and which in 
some modern text-books is called the plane angle of the dihedral angle. 

It is necessary to show that this plane angle is a proper measure of the 
dihedral angle, and accordingly Legendre has a proposition to this effect In 
order to prove it, it is necessary to show that, given two planes meeting in a 
straight line, 

( 1) the plane angle in question is the same at all points of the straight line 
forming the common section ; 

(2) if the dihedral angle between two planes increases or diminishes in a 
certain ratio, the plane angle in question will increase or diminish in the same 
ratio. 

(1) If MAN, MAP be two planes intersecting in MA, and if AN, AP 
be drawn in the planes respectively and at right angles to 
MA, the angle NAP is the inclination of the plane to the 
plane or the plane angle of the dihedral angle. 

Let lifC, MB be also drawn in the respective planes 
at right angles to MA. 

Then since, in the plane MAN, MC and AN are 
drawn at right angles to the same straight line MA, 
MC, AN are parallel. 

For the same reason, MB, AP are parallel. 

Therefore [xt 10] the angle BMC is equal to the 
angle PAN. 

And M may be any point on MA. Therefore the 
plane angle described in the definition is the same at all 
points of AM. 



xl Dzrr. 6— 10] NOTES ON DEFINITIONS s— i° 265 

(a) In the plane A^ P draw the arc NDP of any circle with centre A, 
and draw the radius AD. 

Now the planes NAP, CMB, being both at right angles to the straight 
line MA, are parallel ; [xi, 14] 

therefore the intersections AD, ME of these planes with the plane MAD are 
parallel, [xi. 16] 

and consequently the angles BME, PAD are equal. [xi. 10] 

If now the plane angle NAD were equal to the plane angle DAP, the 
dihedral angle NAMD would be equal to the dihedral angle DAMP; 
for, if the angle PAD were applied to the angle DAN, AM remaining the 
same, the corresponding dihedral angles would coincide. 

Successive applications of this result show that, if the angles NAD, DAP 
each contain a certain angle a certain number of times, the dihedral angles 
NAMD, DAMP will contain the corresponding dihedral angle the same 
number of times respectively. 

Hence, where the angles NAD, DAP are commensurable, the dihedral 
angles corresponding to them are in the same ratio. 

Legendre then extends the proof to the case where the plane angles are 
incommensurable by reference to an exactly similar extension in his proposition 
corresponding to Euclid vi. 1, for which see the note on that proposition. 

Modern text-books make the extension by an appeal to limits. 

Definition 7. 

*E#tVcow Jfpos iriirtiov o/iouut k«A«t&ii \iyvrai not inpov tt/w htpw, S™> 
ai ftprjfiivat tiuv ffXurcuiy yfoviat la at dAA^Aats UKrtp. 

Definition 8. 

1 lapitkkrjkfi {irivcAa itm to. ocrvjUTrTorra. 

Heron has the same definition of parallel planes (Def. r 1 5). The Greek 
word which is translated " which do not meet " is daufHrruircL, the term which 
has been adopted for the asymptotes of a curve. 

Definition 9. 

"O/tota trripia a-^ij/iara hart, Ttt uiro 6/Wajy Iwvx&vnt ircpic;(d/ii*a to- imp to 

TTATJPOS* 

Definition 10. 

*I<ra Se *at opota trrtpta. tr^fund fori ra xnrb afimw nrtircoaiy irfpie^ofjitva 
taav Tip wAiffot tea\ rep fLcyiHtt. 

These definitions, the second of which practically only substitutes the 
words " equal and simitar " for the word " similar " in the first, have been the 
mark of much criticism. 

.Si sn son holds that the equality of solid figures is a thing which ought to be 
primed, by the method of superposition, or otherwise, and hence that Def. 10 
is not a definition but a theorem which ought not to have been placed among 
the definitions. Secondly, he gives an example to show that the definition or 
theorem is not universally true. He takes a pyramid and then erects on the 
base, on opposite sides of it, two equal pyramids smaller than the first. The 
addition and subtraction of these pyramids respectively from the first give two 



*66 BOOK XI [xi. Def. io 

solid figures which satisfy the definition but are clearly not equal (the smaller 
having a re-entrant angle); whence it also appears that two unequal solid 
angles may be contained by the same number of equal plane angles. 

Maintaining then that Def. io is an interpolation by "an unskilful hand," 
Sim son transfers to a place before Def. 9 the definition of a solid angle, and 
then defines similar solid figures as follows : 

Similar solid figures are such as have all their solid angles equal, each to tack, 
and which are contained by (he same number of similar planes. 

Legendre has an invaluable discussion of the whole subject of these 
definitions (Note xn., pp. 3*3 — 336, of the 14th edition of his Aliments de 
Giomitrie). He remarks in the first place that, as Simson said, Def. 10 is not 
properly a definition, but a theorem which it is necessary to prove ; for it is 
not evident that two solids are equal for the sole reason that they have an 
equal number of equal faces, and, if true, the fact should be proved by super- 
position or otherwise. The fault of Def. 10 is also common to Def, 9. For, 
if Def. 1 o is not proved, one might suppose that there exist two unequal and 
dissimilar solids with equal faces ; but, in that case, according to Definition 9, 
a solid having faces similar to those of the two first would be similar to both 
of them, i.e. to two solids of different form : a conclusion implying a con- 
tradiction or at least not according with the natural meaning of the word 
"similar." 

What then is to be said in defence of the two definitions as given by 
Euclid f It is to be observed that the figures which Euclid actually proves 
equal or similar by reference to Deff. 9, io are such that their solid angles do 
not consist of more than three plane angles ; and he proves sufficiently clearly 
that, if three plane angles forming one solid angle be respectively equal to 
three plane angles forming another solid angle, the two solid angles are equal. 
If now two polyhedra have their faces equal respectively, the corresponding 
solid angles will be made up of the same number of plane angles, and the 
plane angles forming each solid angle in one polyhedron will be respectively 
equal to the plane angles forming the corresponding solid angle in tbe other. 
Therefore, if the plane angles in each solid angle are not more than three in 
number, the corresponding solid angles will be equal. But if the correspond- 
ing faces are equal, and the corresponding solid angles equal, the solids must 
be equal ; for they can be superposed, or at least they will be symmetrical 
with one another. Hence the statement of Defl". 9, 10 is true and admissible 
at all events in the case of figures with trihedral angles, which is the only case 
taken by Euclid. 

Again, the example given by Simson to prove the incorrectness of Def. 10 
introduces a solid with a re-entrant angle. But it is more than probable that 
Euclid deliberately intended to exclude such solids and to take cognizance of 
convex polyhedra only ; hence Simson's example is not conclusive against the 
definition. 

Legendre observes that Simson's own definition, though true, has the 
disadvantage that it contains a number of superfluous conditions. To get 
over the difficulties, Legendre himself divides the definition of similar solids 
into two, the first of which defines similar triangular pyramids only, and the 
second (which defines similar polyhedra in general) is based on the first. 

Tifo triangular pyramids are simitar when they have pairs of fates respectively 
similar, similarly placed and equally inclined to one another. 

Then, having formed a triangle with the vertices of three angles taken on 
the same face or base of a polyhedron, we may imagine the vertices of the 



xi. Deit. io, u] NOTES ON DEFINITIONS to, n *6 7 

different solid angles of the polyhedron situated outside of the plane of this 
base to be the vertices of as many triangular pyramids which have the triangle 
for common base, and each of these pyramids will determine the position of 
one solid angle of the polyhedron. This being so, 

Two polyhtdra art similar when they have similar bases, and the vertices of 
their corresponding solid angles outside the bases art defermintd by triangular 
pyramids similar each to each. 

As a matter of fact, Cauchy proved that two convex solid figures are equal 
if they are contained by equal plane figures similarly arranged. Legendre 
gives a proof which, he says, is nearly the same as Cauchy's, depending on two 
lemmas which lead to the theorem that, Givtn a convex polyhedron in which alt 
the solid angles art made up of mort than thret plane angles, it is impossible to 
vary the inclinations of tht plants of this solid so as to product a second polyht- 
dron formed by the same plants arranged in the same manner as in the givtn 
polyhedron. The convex polyhedron in which all the solid angles are made up 
of more than three plane angles is obtained by cutting off from any given 
polyhedron all the triangular pyramids forming trihedral angles (if one and the 
same edge is common to two trihedral angles, only one of these angles is 
suppressed in the first operation). This is legitimate because trihedral angles 
are invariable from their nature. 

Hence it would appear that Heron's definition of equal solid figures, which 
adds " similarly situated " to Euclid's " similar " is correct, if it be understood to 
apply to convex polyhedra only : Equal solid figures are those which are 
contained by equal and similarly situated planes, equal in number and magnitude ; 
where, however, the words " equal and " before " similarly situated " might be 
dispensed with. 

Heron (Def. 118) defines similar solid figures as those which art containtd 
by planes similar and similarly situated. If understood of convex polyhedra, 
there would not appear to be any objection to this, in view of the truth of 
Cauchy's proposition about equal solid figures. 



Definition ii. 

Xttpto. ymvia. itrr'iv i\ vwb ttXiiovwv y St'o ypafi/Atiiv OrWTOfLfvtav aXki'ikivf kol fty 
iv TJj ttvrjj t7ri<f>aytvL owrmv irpov Trafrcus Tats ypafLftais itktGis. *AA\ws ■ trrtptb 
yavia i<rr\v if wro jtVkWc ij %io ymviwv iiiur&tav irtpuxoiiiw) prj ovtruiv cV T<ji 
airr$ brtiri&f. wpos Jet m)/i(iig trvvuJTa.p\ivii>v. 

Heiberg conjectures that the first of these two definitions, which is not in 
Euclid's manner, was perhaps taken by him from some earlier Elements. 

The phraseology of the second definition is exactly that of Plato when he 
is speaking of solid angles in the Timatus (p. 55). Thus he speaks (1) of four 
equilateral triangles so put together {(wttnafitva) that each set of three plane 
angles makes one solid angle, (z) of eight equilateral triangles put together so 
that each set of four plane angles makes one solid angle, and {3) of six squares 
making eight solid angles, each composed of three plane right angles. 

As we know, Apollonius defined an angle as the " bringing together of a 
surface or solid to one point under a broken line or surface." Heron (Def. 12) 
even omits the word " broken " and says that A solid angle is in general (koifw*) 
the bringing together of a surface which has its concavity in one and the samt 
direction to ont point. It is clear from an allusion in Proelus (p. 123, r — 6) to 
the half of a cone cut off by a triangle through the axis, and from a scholium to 



a68 BOOK XI [xi. DerF. n— 13 

this definition, that there was controversy as to the correct ness of describing as a 
solid angle the " angle " enclosed by fewer than three surfaces {including curved 
surfaces). Thus the scholiast says that Euclid's definition of a solid angle as 
made up of three or more plane angles is deficient because it does not e.g. cover 
the case of the angle of a " fourth part of a sphere," which is contained by more 
than two surfaces, though not all plane. But he declines to admit that the 
half-cone forms a solid angle at the vertex, for in that case the vertex of the 
cone would itself be an angle, and a solid angle would then be formed both 
by two surfaces and by one surface; "which is not tme." Heron on the 
other hand (Def. 22) distinctly speaks of solid angles which arc not contained 
by plane rectilineal angles, " e.g. the angles of cones." The conception of the 
latter " angles " as the limit of solid angles with an infinite number of infinitely 
small constituent plane angles does not appear in the Greek geometers so far 
as I know. 

In modern text-books a polyhedral angle is usually spoken of as formed 
(or bounded) by three or more plants meeting at a points or it is the angular 
opening between such planes at the point where they meet. 



Definition 12. 

IIvpa/i« i(m ffJCJ/Ki tmptov fWnrtSow irtpuxa/uroY 6mb ivi/s ivariSou jrpos Ivl 
<rj]jitttf <rw«fr(Js. 

This definition is by no means too clear, nor is the slightly amplified 
definition added to it by Heron (Def. 99). A pyramid is the figure brought 
together to one point, by putting together triangles, from a triangular, quadri- 
lateral or polygonal, that is, any rectilineal, base. 

As we might expect, there is great variety in the definitions given in 
modern text-books. Legendre says a pyramid is the solid formed when several 
triangular planes start from one point and are terminated at the different sides 
of one polygonal plane. 

Mr H. M. Taylor and Smith and Bryant call it a polyhedron all but one of 
whose faces meet in a point. 

Mehler reverses Legendre's form and gives the content of Euclid's in 
clearer language. "An n-sided pyramid is bounded by an n-sided polygon as base 
and n triangUs which connect its sides with one and the same point outside it." 

Rausenberger points out that a pyramid is the figure ci't off from a solid 
angle formed of any number of plane angles by a plane which intersects the 
solid angle. 

Definition 13. 

ITpc'cr/ia tcrrl trxjjfLa trrtptav liFiiri&tH$ irtptt^ofLtvoii, wv Si'o ri Awtvavriov Itra 
T€ Kill ofioia i<m KO.I jrapaXXljka, Ttt Bi Xoilra TrajjaXX^koypaft/ia. 

Mr H. M. Taylor, followed by Smith and Bryant, defines a prism as a 
polyhedron all but two of the faces of which are parallel to one straight line. 

Mehler calls an n-sided prism a body contained between two parallel planes 
and enclosed by n other planes with parallel lines of intersection. 

Heron's definition of a prism is much wider (lief. 105). Prisms are (hose 
figures which are connected \tnmavrorra.) from a rectilineal base to a rectilineal 
area by rectilineal collocation {nar tWvypaniior (rirStatr). By this Heron must 



xi. Dkff. 13—15] NOTES ON DEFINITIONS 11—15 



169 




apparently mean any convex solid formed by connecting the sides and angles 

of two polygons in different planes, and 

each having any number of sides, by 

straight lines forming triangular faces 

{where of course two adjacent triangles 

may be in one plane and so form one 

quadrilateral face) in the manner shown 

in the annexed figure, where ABCD, 

EFG represent the base and its 

opposite. 

Heron goes on to explain that, if 
the face opposite to the base reduces to 
a straight line, and a solid is formed by 
connecting the base to its extremities by 
straight lines, as in the other case, the 
resulting figure is neither a pyramid nor 
a prism. 

Further, he defines parallelogrammic {in the body of the definition parallel- 
sided) prisms as being those prisms which have six faces and have their 
opposite planes parallel. 

Definition 14. 

%$alpd t<JTW, OTtLV TjfJUKVKklav fltvavvr}*; Ttj^ BuLfAtTpOV Tttpiti't^Btv To 

^fUKVukutv tls to avro rdktv diroKanurra^, o0*v rfp£aro tfttpea-Btii, to TrtpiXyftftBh/ 

"MP*. 

The scholiast observes that this definition is not properly a definition 01 a 
sphere but a description of the mode of generating it. But it will be seen, in 
the last propositions of Book xiii., why, Euclid put the definition in this form. 
It is because it is this particular view of a sphere which he uses to prove that 
the vertices of the regular solids which he wishes to " comprehend " in certain 
spheres do lie on the surfaces of those spheres. He proves in fact that the 
said vertices lie on semicircles described on certain diameters of the spheres. For 
the real definition the scholiast refers to Theodosius' Sphaerica. But of course 
the proper definition was given much earlier. In Aristotle the characteristic 
of a sphere is that its extremity is equally distant from its centre (to ttrov Avi^tu/ 
tou fiio-ou to hrxaror, De caelo ii. 14, 297 a 24). Heron (Def. 76) uses the 
same form as that in which Euclid defines the circle : A sphere is a solid 
figure bounded by one surface, such that ail the straight lines falling on it from 
one point of those which lie within the figure are equal to one another. So the 
usual definition in the text-books : A sphere is a closed surface such that all 
points of it are equidistant from a fixed point within it. 



Definition 15. 

*A£u>v &f T199 trfoitpa* itrriv ij jitvovva tvOtia, ir€p\ yv to -rjpXKVttkiov <TTpi<p£T(it. 

That any diameter of a sphere may be called an axis is made clear by 
Heron (Def. 78). The diameter of the sphere is called an axis, and is any 
straight line drawn through the centre and bounded in both directions by the 
sphere, immovable, about which the sphere is moved and turned. Cf. Euclid's 
Def. 17. 



a?° BOOK XI [xi. Dkff. 16—18 

Definition 16. 

KcrTpov Si ttji er^Kupat ta-rl to airo, S xtu Ttru iJjumvKA&u. 

Heron, Def. 77. 7%^ middle {point) of the sphere is called its eentrt ; and 
this same point is also the centre of the hemisphere. 

Definition 17. 

A«£urrpo? Si riff <T$aLpas £<rrtv *v&ta tic Sii tou KtvTpov ify flirty naX ntpa- 
roupirri itf,' inartpa to ft-ipi) wro t^s fari^avitat nj* cr^aifjas. 

Definition 18. 

Kuvac forty, »rav opBoywvLov rptymvov fAtvovtrrfi >uas irXtvpaf Tt3v ir*p* t^v 
op6yy ytm-iiiv Trifittvfx&'ii' to Tplywov <tv to avro iraXip <£irojcaTaaTa$jji, o&tv Tjp$aro 
<f>ipttT&ai, to ir«piXt^fl«i< ireful, xir piv ll fUvowra *v6iw. Iot) 7} Tij) Aoiirfl [171] 
ircol t^v op^v Trtpu$HpQfAfVfl t Gp&o-ytvvtos iarat o kui-os, iav £f MdrrwK, afifikv- 
yuytof, fit Si ^«t£«F, dfuyoSyiot. 

This definition, or rather description of the genesis, of a (right) cone is 
interesting on account of the second sentence distinguishing between right- 
angled, obtuse-angled and acute-angled cones. This distinction is quite 
unnecessary for Euclid's purpose and is not used by him in Book xii. ; it is no 
doubt a relic of the method, still in use in Euclid's time, by which the earlier 
Greek geometers produced conic sections, namely, by cutting right cones only 
by sections always perpendicular to an edge. With this system the parabola 
was a section of a right-angled cone, the hyperbola a section of an obtuse-angled 
cone, and the ellipse a section of an acute-angled cone. The conic sections were 
so called by Archimedes, and generally until Apollonius, who was the first to 
give the complete theory of their generation by means of sections not perpen- 
dicular to an edge, and from cones which are in general oblique circular cones. 
Thus Apollonius begins his Conies with the more scientific definition of a cone. 
If, he says, a straight line infinite in length, and passing always through a fixed 
point, be made to move round the circumference of a circle which is not in the 
same plane with the point, so as to pass successively through every point of 
that circumference, the moving straight line will trace out the surface of a double 
cone, or two similar cones lying in opposite directions and meeting in the fixed 
point, which is the apex of each cone. The circle about which the straight line 
moves is called the base of the cone lying between the said circle and the fixed 
point, and the axis is defined as the straight line drawn from the fixed point, 
or the apex, to the centre of the circle forming the base. Apollonius goes on 
to say that the cone is a scalene or oblique cone except in the particular case 
where the axis is perpendicular to the base. In this latter case it is a right 
cone. 

Archimedes called the right cone an isosceles cone. This fact, coupled 
with the appearance in his treatise On Conoids and Spheroids (7, 8, 9) of 
sections of acute-angled cones (ellipses) as sections of conical surfaces which are 
proved to be oblique circular cones by finding their circular sections, makes it 
sufficiently clear that Archimedes, if he had defined a cone, would have 
defined it in the same way as Apollonius does. 



XI. Deff. ig—iS] NOTES ON DEFINITIONS 16—28 afi 

Definition 19. 

*A£ titV Si TOV KWVOV ItTTtV v) fltYQVfriX tV&ita, Tttpl TJV TO TplytaVQV OTpttjitTUU. 

Definition 20. 

BuiTLS S) 4 X UK A Of i VTTU T7J$ TT f f H<p f flO^I K1J5 IV$(LU<; ypU <f>6fl t V Of 

Definition 21. 

KuAivSpoc iffTik, cTrav op^oyaiWov TrapaAA??\oypd|i^u)u /ifvotlarfs /uaf TrXrvpa? 
T4iV wtpt Tr)k dp^fjk ycjruip Trepupfx&i' to TrapaAArjXoypap^ioy tts to avro iraAt*' 
ijr<wo.T<wrroSB, ofl<c ripiaro aWp«r&u, to irfptXtj^ic o"x^/*a. 

Definition 22, 

Afuf SI tov ttvkty&pav (cTTif tj ji&vavtra. ivStia, iripl 17 k to npaAAiiXdypa/ipoi' 

OTptyfTU. 

Definition 23. 

Bcureie St oi miicAoi ot vto tuk ctirfyaFrcov jrtpuiyofAtvimr £uo irXropfov 

Definition 24. 

'O/ioum jtuj p-oi rat KtiXtrSpot (iatf, wr oT ti uf out ™i al SuVtTpo< r«Sv fidatwr 
iydkoydv &riv. 

Definition 25. 

Ku/9ot furl o-xv/ia o*rtp«S* vjto If wrpaywewe fr™* ir«fHtxo/t(vof . 

Definition 26. 

'OdTatSpiJi' i<m ax*)t"> oTtprev vtto okt4 Tptytuwiiv i<rw ko.i ttroirXtiipojp 

Definition 27. 

EuiOO-OfSpOl' JO-Tt CX^jw <TT(p€OV VTO (IkO^I TplywFMC »0 , UH' KOi Io-twX(VpUH' 

TT<ptf^o/jlivar f 

Definition 28. 

AtaStKatSpoy fori o'x'Jf" 1 or«p«o* fori Sui&ma irtvrayojwitf uruiv Ktti Jo-oirX«iipMK 

KO.L HTOyWVlW TTfplf ^Oft<VOK 







BOOK XI. PROPOSITIONS. 

Proposition i. 

A part of a straight line cannot be in Ike plane of reference 
and a part in a plane more elevated. 

For, if possible, let a part AB of the straight line ABC 
be in the plane of reference, and a part 
BC in a plane more elevated. 

There will then be in the plane of 
reference some straight line continuous 
with AB in a straight line. 

Let it be BD ; 
therefore AB is a common segment of the 
two straight lines ABC, ABD : 
which is impossible, inasmuch as, if we 
describe a circle with centre B and distance 
AB, the diameters will cut off unequal circumferences of the 
circle. 

Therefore a part of a straight line cannot be in the plane 
of reference, and a part in a plane more elevated. 

Q. E. D. 

1. the plane of reference, to irroKtlfitvot {rfacSoy, the plane Inid down or assumed. 

2. more elevated, /LereapoTtptp. 

There is no doubt that the proofs of the first three propositions are 
unsatisfactory owing to the fact that Euclid is not able to make any use of his 
definition of a plane for the purpose of these proofs, and they really depend 
upon truths which can only be assumed as axiomatic. The definition of a plane 
as that surface which lie: evenly with the straight lines on itself, whatever its 
exact meaning may be, is nowhere appealed to as a criterion to show whether 
a particular surface is or is not a plane. If the meaning of it is what I conjec- 
ture in the note on Book ]., Def. 7 (Vol. 1. p. 171), if, namely, it only tries to 
express without an appeal to sight what Plato meant by the " middle covering 
the extremities " (i.e. apparently, in the case of a plane, the fact that a plane 
looked at edgewise takes the form of a straight line), then it is perhaps 
possible to connect the definition with a method of generating a plane which 



xi. i] PROPOSITION r a 73 

has commended itself to many writers as giving a better definition. Thus, if 
we conceive a straight line in space and a point outside it placed so that, in 
Plato's words, the line " covers " the point as we look at them, the line will 
also "cover" every straight line which passes through the given point and 
some one point on the given straight line. Hence, if a straight line passing 
always through a fixed point moves in such a way as to pass successively 
through every point of a given straight line which does not contain the given 
point, the moving straight line describes a surface which satisfies the Euclidean 
definition of a plane as I have interpreted it. But if we adopt the definition 
of a plane as the surface described by a straight line which, passing through a 
given faint, turns about it in such a -way as always to intersect a given straight 
line not passing through the given point, this definition, though it would help us 
to prove Eucl. *Xi. 2, does not give us the fundamental properties of a plane ; 
some postulate is necessary in addition. The same is true even if we take a 
definition which gives more than is required to determine a plane, the defini- 
tion known as Simson's, though it is at least as early as the time of Theon of 
Smyrna, who says (p. 112,5) tnat a P^ nt ** a surface such that, if a straight line 
meet it in two points, the straight line lies wholly in it (oAij avrif i^oppifrrat). 
This is also called the axiom of the plane. (For some attempts to prove this on 
the basis of other definitions of a plane see my note on the definition of a plane 
surface, t. Def. 7.) If this definition or axiom be assumed. Prop. 1 becomes 
evident, for, as I,egendre says, "In accordance with the definition of the plane, 
when a straight line has two points common with a plane, it lies wholly in the 
plane." 

Euclid practically assumes the axiom when he says in this proposition 
"there will be in the plane of reference some straight line continuous with 
AB" Clavius tries, unsuccessfully, to deduce this from Euclid's own 
definition of a plane ; and he seems to admit his 
failure, because he proceeds to try another tack. 
Draw, he says, in the plane DE, the straight tine 
CG at right angles to AC, and, again in the plane 
DE, CF at right angles to CG (t. 1 r]. Then AC, 
CF make right angles with CG in the same plane ; 
therefore (1. 14) ACFis a straight line. But this 
does not really help, because Euclid assumes tacitly, 
in Book i. as well as Book xi., that a straight line joining two points in a 
plane lies wholly in that plane. 

A curious point in Euclid's proof is the reason given why two straight lines 
cannot have a common segment. The argument is precisely that of the 
" proof" of the same thing given by Proclus on 1. 1 (see note on Book 1. 
Post J, Vol. 1. p. r97) and is of course inconclusive. The fact that two 
straight lines cannot have a common segment must be taken to be involved 
in the definition of, and the postulates relating to, the straight line; and the 
" proof" given here can hardly, I should say, be Euclid's, though the interpo- 
lation, if it be such, must have been made very early. 

The proof assumes too that a circle can be dpseribed so as to cut BA, BC 
and SB, or, in other words, it assumes that AD, BC are in one plane; -that 
is, Prop. 1 as we have it realty assumes the result of Prop. 2. There is there- 
fore ground for Simson's alteration of the proof (after the point where 3D has 
been taken in the given plane in a straight line with AB) to the following : 

" Let any plane pass through the straight line AD and be turned about it 
until it pass through the point C 




a74 BOOK XI [xi. i, 2 

And, because the points B, C are in this plane, the straight line BC is 
in it. [Sim son's def.] 

Therefore there are two straight lines ABC, ABD in the same plane that 
have a common segment AB : 
which is impossible." 

Simson, of course, justifies the last inference by reference to his Corollary 
to i. i r, which, however, as we have seen, is not a valid proof of the assump- 
tion, which is really implied in i. Post. 2. 

An alternative reading, perhaps due to Theon, says, after the words 
"which is impossible " in the Greek text, "for a straight line does not meet a 
straight line in more points than one ; otherwise the straight lines will 
coincide." Simson (who however does not seem to have had the second 
clause beginning "otherwise" in the text which he used) attacks this alterna- 
tive reading in a rather confused note chiefly directed against a criticism by 
Thomas Simpson, without (as it seems to me) sufficient reason. It contains 
surely a legitimate argument. The supposed straight lines ABC, ABD meet 
in more than two points, namely in alt the points between A and B. But two 
straight lines cannot have two points common without coinciding altogether ; 
therefore ABC must coincide with ABD. 

Proposition 2. 

If two straight lines cut one another, they are in one plane, 
and every triangle is in on* plane. 

For let the two straight lines AB, CD cut one another at 
the point £ ; 

I say that AB, CD are in one plane, 
and every triangle is in one plane. 

For let points F, G be taken at 
random on EC, EB, 
let CB, FG be joined, 
and let FH, GK be drawn across ; 
I say first that the triangle ECB is 
in one plane. 

For, if part of the- triangle ECB, 
either FHC or GBK, is in the plane of reference, and the rest 
in another, 

a part also of one of the straight lines EC, EB will be in the 
plane of reference, and a part in another. 

But, if the part FCBG of the triangle ECB be in the 
plane of reference, and the rest in another, 
a part also of both the straight lines EC, EB will be in the 
plane of reference and a part in another : 
which was proved absurd. [su. 1] 




xi. 2] PROPOSITIONS i, 2 175 

Therefore the triangle ECB is in one plane. 
But, in whatever plane the triangle ECB is, in that plane 
also is each of the straight lines EC, EB, 

and, in whatever plane each of the straight lines EC, EB is, 
in that plane are AB, CD also. [xt. 1] 

Therefore the straight lines AB, CD are in one plane, 

and every triangle is in one plane. 

Q. E. D. 

It must be admitted that the " proof" of this proposition is not of any 
value. For one thing, Euclid only takes certain triangles and a certain 
quadrilateral respectively forming part of the original triangle, and argues 
about these. But, for anything we are supposed to know, there may be some 
part of 'hi; triangle bounded (let us say) by some curve which is not in the 
same plane with the triangle. 

We may agree with Simson that it would be preferable to enunciate the 
proposition as follows. 

Tiw straight lints which intersect are in one plane, and three straight lines 
which intersect two and two are in one plane. 

Adopting Smith and Bryant's figure in preference to Sim son's, we suppose 
three straight lines PQ, RS, XYto intersect 
two and two in A , B, C. R p 

Then Simson's proof (adopted by Legen- \*S 

dre also) proceeds thus. Vf 

Let any plane pass through the straight S\ 
line PQ, and let this plane be turned about / \ 
PQ (produced indefinitely) as axis until it / \ 
passes through the point C. x / \ y 

Then, since the points A, C are in this ~7^ "\ 

plane, the straight line AC (and therefore / \ 

the straight tine RS produced indefinitely) *• Jj 

lies wholly in the plane. [Simson's def.j 

For the same reason, since the points />', C are in the plane, the straight 
line XYMes wholly in the plane. 

Hence all three straight lines PQ, PS, XY (and of course any pair of 
them) lie in one plane. 

But it has still to be proved that there is only one plane passing through 
the three straight lines. 

This may be done, as in Mr Taylor's Euclid, thus. 

Suppose, if possible, that there are two different planes through A, B, C. 

The straight lines BC, CA, AB then lie wholly in each of the two planes. 

Now any straight line in one of the two planes must intersect at least two 
of the straight lines (produced if necessary) ; 

let it intersect two of them in K, L. 

Then, since K, L are also in the second plane, the line KL lies wholly in 
that plane. 

Hence every straight line in either of the planes lies wholly in the other 
also ; and therefore the planes are coincident throughout their whole surface. 



876 BOOK XI [». a, 3 

It follows from the above that 

A plan/ is determined (i.e. uniquely determined} by any of the following data : 
(1) by thru straight lines meeting one another two and two, 
(t) by three points not in a straight line, 

(3) by tivo straight lines meeting one another, 

(4) by a straight line and a point without it. 

Proposition 3. 

If two planes cut one another, their common section is a 
straight tine. 

For let the two planes AB, BC cut one another, 
and let the line DB be their common 
section ; 
I say that the line DB is a straight line. 

For, if not, from D to B let the straight 
line DEB be joined in the plane AB, and 
in the plane BC the straight line DFB. 

Then the two straight lines DEB, DFB 
will have the same extremities, and will 
clearly enclose an area : 
which is absurd. 

Therefore DEB, DFB are not straight lines. 

Similarly we can prove that neither will there be any 
other straight line joined from DtoB except DB the common 
section of the planes AB, BC. 

Therefore etc. 

Q. E. D. 

I think Simson is right in objecting Co the words after " which is absurd/' 
to the effect that DEB, DFB are not straight tines, and that neither can there 
be any other straight line joined from D to B except DB, as being unneces- 
sary. It is right to conclude at once from the absurdity that 3D cannot but 
be a straight line. 

Legendre makes his proof depend on Prop. 2. " For, if, among the points 
common to the two planes, three should be found which are not in a straight 
line, the two planes in question, each passing through three points, would only 
amount to one and the same plane." [This of course assumes that three 
points determine one and only one plane, which, strictly speaking, involves 
more than Prop. a itself, as shown in the last note.] 

A favourite proposition in modern text-books is the following. The proof 
seems to be due to von Staudt (Killing, Grundlagen der Geometrie, Vol. 11. 
P- «)■ 




xi. 3. 4] 



PROPOSITIONS 3—4 



377 



J] two plants meet in a point, they meet in a straight line. 

Let ABC, ADE be two given planes meeting 
at A. 

Take any points B, C lying on the plane ABC, 
and not on the plane ADE but on the same side 
of it 

Join AB, AC, and produce BA to F. 

Join CF. 

Then, since B, Fare on opposite sides of the 
plane ADE, 
C, Fare also on opposite sides of it. 

Therefore CF must meet the plane ADE in 
some point, say G. 

Then, since A, G are both in each of the planes ABC, ADE, the straight 
line AG is in both planes. [Simson's def.J 

This is also the place to insert the proposition that, // three planes intersect 
two and two, their lines of intersection either meet in a point or are parallel two 
and two. 

Let there be three planes intersecting in the straight lines AB, CD, EF. 




cce? 







Now AB, EF are in a plane ; therefore they either meet in a point or are 
parallel. 
(i) Let them meet in O. 

Then O, being a point in AB, lies in the plane AD, and, being also a 
point in EF, lies also in the plane ED. 

Therefore O, being common to the planes AD, DE, must lie on CD, the 
line of their intersection ; 
i.e. CD, if produced, passes through O, 
(3) Let AB, EF not meet, but let them be parallel. 

Then CD cannot meet AB ; for, if it did, it must necessarily meet EF, 
by the first case. 

Therefore CD, AB, being in one plane, are parallel. 

Similarly CD, EF&te parallel. 

Proposition 4. 

If a straight line be set up at right angles to two straight 
lines which cut one another, at their common point of section, 
it will also lie at right angles to the plane through them. 




*78 BOOK XI [xi. 4 

For let a straight line EF be set up at right angles to the 
two straight lines AB, CD, which 
cut one another at the point E, 
from E ; 

I say that EF is also at right 
angles to the plane through AB, 
CD. 

For let AE, EB, CE, ED be 
cut off equal to one another, 

and let any straight line GEH be drawn across through E, 
at random ; 
let AD, CB be joined, 

and further let FA, FG, FD, FC, FH, FB be joined from 
the point F taken at random <on FF>. 

Now, since the two straight lines AE, ED are equal to 
the two straight lines CE, EB, and contain equal angles, [i. 15] 
therefore the base AD is equal to the base CB, 
and the triangle AED will be equal to the triangle CEB; [1.4] 
so that the angle DAE is also equal to the angle EBC. 

But the angle AEG is also equal to the angle BEH ;[i. 15] 
therefore AGE, BEH are two triangles which have two 
angles equal to two angles respectively, and one side equal 
to one side, namely that adjacent to the equal angles, that 
is to say, AE to EB ; 

therefore they will also have the remaining sides equal to the 
remaining sides. ['• 26] 

Therefore GE is equal to EH, and AG to BH. 

And, since AE is equal to EB, 
while FE is common and at right angles, 
therefore the base FA is equal to the base FB. [1. 4] 

For the same reason 
FC is also equal to FD. 

And, since AD is equal to CB, 
and FA is also equal to FB, 

the two sides FA, AD are equal to the two sides FB, BC 
respectively ; 

and the base FD was proved equal to the base FC ; 
therefore the angle FAD is also equal to the angle FBC. [1. 8] 



xi. 4] PROPOSITION 4 j 79 

And since, again, AG was proved equal to BH, 
and further FA also equal to FB, 
the two sides FA, AG are equal to the two sides FB BH. 

And the angle FAG was proved equal to the angle FBH; 
therefore the base FG is equal to the base FH. [i. 4] 

Now since, again, GE was proved equal to EH, 
and EF is common, 

the two sides GE, EFare equal to the two sides HE, EF; 
and the base FG is equal to the base FH; 
therefore the angle GEF is equal to the angle HEF. [1. 8] 

Therefore each of the angles GEF, HEF is right. 

Therefore FE is at right angles to GH drawn at random 
through E. 

Similarly we can prove that FE will also make right 
angles with all the straight lines which meet it and are in the 
plane of reference. 

But a straight line is at right angles to a plane when it 
makes right angles with all the straight lines which meet it 
and are in that same plane ; [xi. Def. 3] 

therefore FE is at right angles to the plane of reference. 

But the plane of reference is the plane through the straight 
lines AB, CD. 

Therefore FE is at right angles to the plane through 
AB, CD. 

Therefore etc. 

Q. E. D. 

The steps to be successively proved in order to establish this proposition 
by Euclid's method are 

(1) triangles A ED, EEC equal in all respects, [by 1. 4] 

(t) triangles AEG, BE/fequ&l in all respects, [by 1. 26] 

so that AG is equal to BH, and GE to EH, 

(3) triangles AEF, BEF equal in all respects, [1. 4] 
so that AFis equal to BE, 

(4) likewise triangles CEF, DEF, 
so that CFis equal to DF, 

(5) triangles FAD, FBC equal in all respects, [1. 8] 
so that the angles FAG, FBH&m equal, 

(6) triangles FAG, FBH equal in all respects, [by (1), (3), {5) and 1. 4] 
so that FG is equal to FH, 



a8o 



BOOK XI 



[xi. 4 




(7) triangles FEG, FEH equal in all respects, [by (2), (6) and 1. 8] 

so that the angles FEG, FEH axe equal, 
and therefore FE is at right angles to Gil. 

In consequence of the length of the above proof others have been 
suggested, and the proof which now finds most general acceptance is that of 
Cauchy, which is as follows. 

Let AB be perpendicular to two straight lines BC, BD in the plane MN 
at their point of intersection B. 

In the plane vlWdraw BE, any straight line 
through B. 

Join CD, and let CD meet BE in ML 

Produce ABto F%o that BF is equal to AB, 

Join AC, AE, AD, CF, EF, DF. 

Since BC is perpendicular to AF at its 
middle point B, 
AC is equal to CF. 

Similarly AD is equal to DF. 

Since in the triangles A CD, FCD the two 
sides AC, CD are respectively equal to the two 
sides FC, CD, and the third sides AD, ED are 
also equal, 

the angles ACD, FCD are equal. [i. 8] 

The triangles ACE, FCE thus have two sides and the included angle 
equal, whence 

EA is equal to EF. [1. 4] 

The triangles ABE, FBE have now all their sides equal respectively ; 

therefore the angles ABE, FBE are equal, [1. 8] 

and AB is perpendicular to BE. 

And BE is in any straight line through B in the plane MN. 

Legendre's proof is not so easy, but it is interesting. We are first required 
to draw through any point E within the angle 
CBD a straight line CD bisected at E. 

To do this we draw EK parallel to DB 
meeting B C in K, and then mark off KC equal 
to BK. 

CE is then joined and produced to D; and 
CD is the straight line required. 

Now, joining AC, AE, AD in the figure 
above, we have, since CD is bisected at E, 
(i) in the triangle ACD, 

AC + AD 1 =2AE i +2ED', 
and also (2) in the triangle BCD, 

BC + BDP = 2BF? + tED 1 . 

Subtracting, and remembering that the triangles ABC, ABD are right- 
angled, so that 

AC'-BC^AB*, 
and AD t -BD> = AB l , 

we have sAB? = 2AE> - 2BE 1 , 

or AE'^AB' + BE 1 , 




xi. 4, 5] PROPOSITIONS 4, 5 181 

whence [1. 48] the angle ABE is a right angle, and AB is perpendicular 
to BE. 

It follows of course from this proposition that the perpendicular AB is the 
shortest distant from A to the plant MN. 

And it can readily be proved that, 

If f rem a point without a plane oblique straight lines be drawn to the plane, 
( i ) those meeting the plane at equal distances from the foot of the perpendicular 
art equal, and 

(i) of ttvo straight lines meeting the plane at unequal distances from the foot of 
the perpendicular, the more remote is the greater. 

Lastly, it is easily seen that 

From a point outside a plane only one perpendicular can be drawn to that 
plane. 

For, if possible, let there be two perpendiculars. Then a plane can be 
drawn through them, and this will cut the original plane in a straight line. 

This straight line and the two perpendiculars will form a plane triangle 
which has two right angles : which is impossible. 



Proposition 5. 

If a straight line be set up at right angles to three straight 
lines which meet one another, at their common point of section, 
the three straight lines are in one plane. 

For let a straight line AB be set up at right angles to the 
three straight lines BC, BD, BE, at 
their point of meeting at B ; 
I say that BC, BD, BE are in one plane. 

For suppose they are not, but, if 
possible, let BD, BE be in the plane of 
reference and BC in one more elevated ; 
let the plane through AB, BC be 
produced ; 

it will thus make, as common section in the plane of reference, 
a straight line. [xi. 3] 

Let it make BF. 

Therefore the three straight lines AB, BC, BE are in one 
plane, namely that drawn through AB, BC. 

Now, since AB is at right angles to each of the straight 
lines BD, BE, 

therefore AB is also at right angles to the plane through 
BD, BE. [xi. 4] 




282 



BOOK XI 



[x'-5 



But the plane through BD, BE is the plane of reference ; 
therefore AB is at right angles to the plane of reference. 

Thus AB will also make right angles with all the straight 
lines which meet it and are in the plane of reference. 

[xi. De£ 3] 

But BF which is in the plane of reference meets it ; 

therefore the angle ABF is right. 

But, by hypothesis, the angle ABC is also right ; 
therefore the angle ABF is equal to the angle ABC. 

And they are in one plane : 
which is impossible. 

Therefore the straight line BC is not in a more elevated 
plane ; 

therefore the three straight lines BC BD, BE are in one 
plane. 

Therefore, if a straight line be set up at right angles to 
three straight lines, at their point of meeting, the three straight 
lines are in one plane. Q. e. d. 

It follows that, // a right angle be turned about one of the straight lines 
containing it the other will describe a plane. 

At any point in a straight line it is possible to draw only one plane which 
is at right angles to the straight line. 

One such plane can be found by taking any two planes through the given 
straight line, drawing perpendiculars to the straight 
line in the respective planes, e.g. BO, CO in the 
planes AOB, AOC, each perpendicular to AO, 
and then drawing a plane (BOC) through the 
perpendiculars. 

If there were another plane through O per- 
pendicular to AO, it must meet the plane through 
AO and some perpendicular to it as OC in a 
straight line OC different from OC. 

Then, by xi. 4, AOC is a right angle, and in 
the same plane with the right angle AOC ■■ which is impossible. 

Next, one plane and only one can be drawn through a point outside a straight 
line at right angUs to that line. 

Let P be the given point, AB the given straight 
line. 

In the plane through P and AB, draw PO per- 
pendicular to AB, and through draw another straight 
line OQ at right angles to AB. 

Then the plane through OP, OQ is perpendicular 
to AB. 

If there were another plane through P perpendicular 
to AB, either 





xi. 5. 6] PROPOSITIONS 5, 6 383 

<i) it would intersect AB at O but not pass through OQ, or 
(2) it would intersect AB At a point different from Q. 



In either case, an absurdity would result 




Proposition 6. 

If two straight lines be at right angles to the same plane, 
the straight lines ■mill be parallel. 

For let the two straight lines AB, CD be at right angles 
to the plane of reference ; 
I say that AB is parallel to CD. 

For let them meet the plane of 
reference at the points B, D, 
let the straight line BD be joined, 
let DE be drawn, in the plane of 
reference, at right angles to BD, 
let DE be made equal to AB, 
and let BE, AE, AD be joined. 

Now, since AB is at right angles to the plane of reference, 
it will also make right angles with all the straight lines which 
meet it and are in the plane of reference. [xi. Def. 3] 

But each of the straight lines BD, BE is in the plane of 
reference and meets AB; 
therefore each of the angles ABD t ABE is right. 

For the same reason 
each of the angles CDB, CDS is also right. 

And, since AB is equal to DE, 
and BD is common, 

the two sides AB, BD are equal to the two sides ED, DB ; 
and they include right angles ; 
therefore the base AD is equal to the base BE, [1. 4] 

And, since AB is equal to DE, 
while AD is also equal to BE, 

the two sides AB, BE are equal to the two sides ED, DA ; 
and AE is their common base ; 
therefore the angle ABE is equal to the angle EDA. [1. 8] 



184 



BOOK XI 



[xi. 6 



But the angle ABE is right ; 
therefore the angle EDA is also right ; 
therefore ED is at right angles to DA. 

But it is also at right angles to each of the straight lines 
BD, DC; 

therefore ED is set up at right angles to the three straight 
lines BD, DA, DC at their point of meeting; 
therefore the three straight lines BD, DA, DC are in one 
plane. [xi. 5] 

But, in whatever plane DB, DA are, in that plane is AB 
also, 

for every triangle is in one plane ; [xi. *] 

therefore the straight lines AB, BD t DC are in one plane 

And each of the angles ABD, BDC is right ; 
therefore AB is parallel to CD. [1. *&\ 



Therefore etc. 



Q. E, D. 



If anyone wishes to convince himself of the real necessity for some 
general agreement as to the order in which propositions in elementary 
geometry should be taken, let him contemplate the hopeless result of too 
much independence on the part of editors in the matter of this proposition 
and its converse, xi. 8. 

Legendre adopts a different, and elegant, method of proof ; but he applies 
it to xi. 8, which he gives first, and then deduces xt. 6 from it by rtductio ad 
absurdum. Dr Mehler uses Legendre's method of proof but applies it to 
xi. 6, and then gives xi. 8 as a deduction from it I^ardner follows Legendre. 
Holgate, the editor of a recent American book, gives Euclid's proof of XL 6 
and deduces xi. 8 by reduetio ad absurdum. Mis countrymen, Schultze and 
Sevenoak, give xi, 8 first, but put it after, and deduce it from, Eucl. xi. 10; 
they then give xi. 6, practically as a deduction from XL 8 by reduetio ad 
absurdum, after a proposition corresponding to Eucl. xi. ti and Iff, and a 
corollary to the effect that through a given point one and only one perpen- 
dicular can be drawn to a given plane. 

We will now give the proof of xi. 6 by Legendre's method {adopted by 
Siiith and Bryant as well as by Mehler). 

Let AB, CD be both perpendicular to the 
same plane MN. 

Join BD. 

Now, since BD meets AB, CD, both of 
which are perpendicular to the plane MN in 
which BD is, 
the angles ABD, CDS are right angles, 

AB, CD will therefore be parallel provided 
that they are in the same plane. 

Through D draw EDF, in the plane MN, 
at right angles to BD, and make ED equal to DF. 







xi. 6, 7 J PROPOSITIONS 6, 7 385 

Join BE, BF, AE, AD, AF. 

Then the triangles BDE, BDF are equal in all respects (by 1. 4), so that 
BE is equal to BF, 

It follows, since the angles ABE, ABFaxe right, that the triangles ABE, 
ABFaie equal in all respects, and 

AE is equal to AF. 

[Mehler now argues elegantly thus. If C£, CF be also joined, it is clear 
that 

CE is equal to CF. 

Hence each of the four points A, B, C, D is equidistant from the two 
points E, F. 

Therefore the points A, B, C, D are in one plane, so that AB, CD are 
parallel. 

If, however, we do not use the locus of points equidistant from two fixed 
points, we proceed as follows.] 

The triangles A ED, AFD have their sides equal respectively j 
hence [1. 8] the angles ADE, ADF aje equal, 
so that ED is at right angles to AD. 

Thus ED is at right angles to BD, AD, CD; 
therefore CD is in the plane through AD, BD. [xi. 5] 

But AB is in that same plane; [xi. z] 

therefore AB, CD are in the same plane. 

And the angles ABD, CDB are right ; 
therefore AB, CD are parallel. 

Proposition 7. 

If two straight lines be parallel and points be taken at 
random on each of them, the straight line joining the points is 
in the same plane with the parallel straight lines. 

Let AB, CD be two parallel straight lines, 
and let points E, F be taken at random 

on them respectively ; 

I say that the straight line joining the 
points E, F is in the same plane with 
the parallel straight lines. 

For suppose it is not, but, if possible, c 
let it be in a more elevated plane as 
EGF, 

and let a plane be drawn through EGF; 
it will then make, as section in the plane of reference, a 
straight line. [xi. 3] 



a86 BOOK XI [xi. 7 

Let it make it, as EF; 
therefore the two straight lines EGF, EF will enclose an 
area ; 
which is impossible. 

Therefore the straight line joined from E to F is not in a 
plane more elevated ; 

therefore the straight line joined from E to F is in the plane 
through the parallel straight lines AB, CD. 

Therefore etc. 

Q. E. D, 

It is true that this proposition, in the form in which Euclid enunciates it, 
is hardly necessary if the plane is defined as a surface such that, if any two 
points be taken in it, the straight line joining them lies wholly in the surface. 
But Euclid did not give this definition ; and, moreover, Prop. 2 would be 
usefully supplemented by a proposition which should prove that two parallel 
straight lines determine a plane (i.e. one plane and one only) which also 
contains alt the straight lines which join a point on one of the parallels to a point 
on the other. That there cannot be two planes through a pair of parallels 
would be proved in the same way as we prove that two or three intersecting 
straight lines cannot be in two different planes, inasmuch as each transversal 
lying in one of the two supposed planes through the parallels would lie wholly 
in the other also, so that the two supposed planes must coincide throughout 
(cf. note on Prop. 2 above). 

But, whatever be' the value of the proposition as it is, Simson seems to 
have spoilt it completely. He leaves out the construction of a plane through 
EGF, which, as Euclid says, must cut the plane containing the parallels in 
a straight line ; and, instead, he says, " In the plane ABCD in which the 
parallels are draw the straight line EHF from £ to F." Now, although we 
can easily draw a straight line from E to F, to claim that we can draw it in 
the plane in which the parallels are is surely to assume the very result which is 
to be proved. All that we could properly say is that the straight line joining 
E to F is in some plane which contains the parallels ; we do not know that 
there is no more than one such plane, or that the parallels determine a plane 
uniquely \ without some such argument as that which Euclid gives. 

Nor can I subscribe to the remarks in Simson 's note on the proposition. 
He says (i) "This proposition has been put into this book by some unskilful 
editor, as is evident from this, that straight lines which are drawn from one 
point to another in a plane are, in the preceding books, supposed to be in that 
plane ; and if they were not, some demonstrations in which one straight line 
is supposed to meet another would not be conclusive. For instance, in 
Prop. 30, Book 1, the straight line GK would not meet EF, if GK were not in 
the plane in which are the parallels AB, CD, and in which, by hypothesis, the 
straight line EF is." But the subject-matter of Book t. and Book xi. is quite 
different ; in Book 1. everything is in one plane, and when Euclid, in defining 
parallels, says they are straight lines in the same plant etc., he only does so 
because he must, in order to exclude non-intersecting straight lines which are 
not parallel. Thus in 1. 30 there is nothing wrong in assuming that there may 
be three parallels in one plane, and that the straight line GHK cuts all three. 



xi. 7, 8] PROPOSITIONS 7, 8 287 

But in Book xi. it becomes a question whether there can be more than one 
plane through parallel straight lines. 

Simson goes on to say (2) "Besides, this 7th Proposition is demonstrated 
by the preceding 3rd ; in which the very same thing which is proposed to be 
demonstrated in the 7th is twice assumed, vi^., that the straight line drawn 
from one point to another in a plane is in that plane." But there is nothing 
in Prop. 3 about a plane in which two parallel straight lines are ; therefore 
there is no assumption of the result of Prop. 7. What is assumed is that, 
given two points in a plane, they can be joined by a straight line in the plane : 
a legitimate assumption. 

Lastly, says Simson, "And the same thing is assumed in the preceding 
6th Prop, in which the straight line which joins the points B, D that are in 
the plane to which AB and CD are at right angles is supposed to be in that 
plane." Here again there is no question of a plane in which two parallels are ; 
so that the criticism here, as with reference to Prop, 3, appears to rest on a 
misapprehension. 



Proposition 8. 

If two straight tines be parallel, and one of them be at 
right angles to any plane, the remaining one will also be at 
right angles to ike same plane. 

Let AB, CD be two parallel straight lines, 
and let one of them, AB, be at right 
angles to the plane of reference ; 
I say that the remaining one, CD, will 
also be at right angles to the same 
plane. 

For let AB, CD meet the plane of 
reference at the points B, D, 
and let BD be joined ; 
therefore AB, CD, BD are in one plane. [xi 7] 

Let DE be drawn, in the plane of reference, at right angles 
to BD, 

let DE be made equal to AB, 
and let BE, AE, AD be joined. 

Now, since AB is at right angles to the plane of reference, 
therefore AB is also at right angles to all the straight lines 
which meet it and are in the plane of reference ; [xi. Def. 3] 
therefore each of the angles ABD, ABE is right. 

And, since the straight line BD has fallen on the parallels 
AB, CD, 




*88 BOOK XI [xi. 8 

therefore the angles ABD, CDB are equal to two right 
angles. [■■ 29] 

But the angle ABD is right ; 
therefore the angle CDB is also right ; 
therefore CD is at right angles to BD. 

And, since AB is equal to DE, 
and BD is common, 

the two sides AB, BD are equal to the two sides ED, DB ; 
and the angle ABD is equal to the angle EDB, 
for each is right ; 
therefore the base AD is equal to the base BE, 

And, since AB is equal to DE, 
and BE to AD, 

the two sides AB, BE are equal to the two sides ED, DA 
respectively, 

and AE is their common base ; 
therefore the angle ABE is equal to the angle EDA. 

But the angle ABE is right ; 
therefore the angle EDA is also right ; 
therefore ED is at right angles to AD. 

But it is also at right angles to DB ; 
therefore ED is also at right angles to the plane through 
BD, DA, [xi. 4] 

Therefore ED will also make right angles with all the 
straight lines which meet it and are in the plane through 
BD, DA. 

But DC is in the plane through BD, DA, inasmuch as 
AB, BD are in the plane through BD, DA, [xi. 1] 

and DC is also in the plane in which AB, BD are. 

Therefore ED is at right angles to DC, 
so that CD is also at right angles to DE. 

But CD is also at right angles to BD. 

Therefore CD is set up at right angles to the two straight 
lines DE, DB which cut one another, from the point of section 
atD; 



xi. 8] 



PROPOSITION 8 



*8o 



so that CD is also at right angles to the plane through 
DE, DB. [xi. 4 ] 

But the plane through DE, DB is the plane of reference ; 
therefore CD is at right angles to the plane of reference. 

Therefore etc. 

Q. E. D. 

Simson objects to the words which explain why DC is in the plane through 
BD, DA, viz. "inasmuch as AB, BD are in the plane through BD, DA, and 
DC is also in the plane in which AB, BD are," as being too roundabout. 
He concludes that they are corrupt or interpolated, and that we ought only to 
have the words " because all three are in the plane in which are the parallels 
AB, CD " (by Prop. 7 preceding). But I think Euclid's words can be 
defended. Prop. 7 says nothing of a plane determined by two transversals as 
BD, DA are. Hence it is natural to say that DC is in the same plane in 
which AB, BD are [Prop. j\ and AB, BD are in the same plane as BD, 
DA [Prop, 3], so that DC is in the plane through BD, DA. 

Legend re s alternative proof is split by him into two propositions. 

(1) Let AB be a perpendicular to (heptane MN and EF a tine situated in that 
piane ; if from B, the foot of the perpendicular, BD be drawn perpendicular to 
EF, and AD be joined, I say that AD will be perpendicular to EF. 

(2) If AB is perpendicular to the plane MN, every straight line CD parallel to 
AB will be perpendicular to the same plane. 

To prove both propositions together we suppose CD given, join BD, 
and draw EF perpendicular to BD in the 
plane MN. 

(t) As before, we make DE equal to DEand 
join BE, BE, AE, AF. 

Then, since the angles BDE, BDF are 
right, and DE, DF equal, 

BE is equal to BF. [1. 4] 

And, since AB is perpendicular to the 
plane, 

the angles ABE, ABE axe both right. 
Therefore, in the triangles ABE, ABF, 

AE is equal to AF. [1. 4] 

Lastly, in the triangles ADE, ADF, since AE is equal to AF, and DE 
to DF, while AD is common, 

the angle ADE is equal to the angle ADF, [1. 3] 

so that AD is perpendicular to EF, 

(2) ED being thus perpendicular to DA, and also (by construction) 
perpendicular to DB, 

ED is perpendicular to the plane ADB. [xi. 4] 

But CD, being parallel to AB, is in the plane ABD ; 

therefore ED is perpendicular to CD. [xt. Def. 3] 




89° 



BOOK XL [xt. 8, 9 




Also, since AB, CD are parallel, 
and ABD is a right angle, 
CDB is also a right angle. 

Thus CD is perpendicular to both DE and DB, and therefore to the 
plane MN through DE, DB. 



Proposition g. 

Straight lines which are parallel to the same straight line 
and are not in the same plane with it are also parallel to one 
another. 

For let each of the straight lines AB, CD be parallel to 
EF, not being in the same plane 

with it ; b h * 

I say that AB is parallel to CD. 

For let a point G be taken at f Q^ e 

random on EF, 

and from it let there be drawn 

GH, in the plane through EF, 

AB, at right angles to EF, and GK in the plane through 

FE, CD again at right angles to EF. 

Now, since EF is at right angles to each of the straight 
lines GH, GK, 

therefore EF is also at right angles to the plane through 

GH, GK. [xi. 4 ] 

And EF is parallel to AB ; 

therefore AB is also at right angles to the plane through 

HG, GK. [xi. 8] 

For the same reason 
CD is also at right angles to the plane through HG, GK ; 
therefore each of the straight lines AB, CD is at right angles 
to the plane through HG,. GK. 

But, if two straight lines be at right angles to the same 
plane, the straight lines are parallel ; [xi. 6] 

therefore AB is parallel to CD. 



xi. 10] PROPOSITIONS 8— 10 291 

Proposition 10. 

If two straight lines meeting one another be parallel to 
two straight lines meeting one another not in the same plane, 
they will contain equal angles. 

For let the two straight lines AB, BC meeting one 
another be parallel to the two straight lines DE, EF meeting 
one another, not in the same plane ; 
I say that the angle ABC is equal to the angle DBF, 




For let BA, BC, ED, EF be cut off equal to one another, 
and let AD, CF, BE, AC, DF be joined. 

Now, since BA is equal and parallel to ED, 
therefore AD is also equal and parallel to BE. [1. 33] 

For the same reason 
CF is also equal and parallel to BE. 

Therefore each of the straight lines AD, CF is equal and 
parallel to BE. 

But straight lines which are parallel to the same straight 
line and are not in the same plane with it are parallel to one 
another ; [xi. 9] 

therefore AD is parallel and equal to CF. 

And AC, DF join them ; 
therefore AC is also equal and. parallel to DF. [1. 33] 

Now, since the two sides AB, BC are equal to the two 
sides DE, EF, 

and the base AC is equal to the base DF, 
therefore the angle ABC is equal to the angle DEF. [1. 8] 

Therefore etc. 



292 BOOK XI [xi. 10, ii 

The result of this proposition does not appear to be quoted in Euclid until 
xii. 3; but Euclsd no doubt inserted it here advisedly, because it has the 
effect of incidentally proving that the "inclination of two planes to one 
another," as defined in XL Def. 6, is one and the same angle at whatever 
point of the common section the plane angle measuring it is drawn. 










Proposition ii. 

Front a given elevated point to draw a straight line perpen- 
dicular to a given plane. 

Let A be the given elevated point, and the plane of 
reference the given plane ; 
thus it is required to draw from the 
points a straight line perpendicular to 
the plane of reference. 

Let any straight line BC be drawn, 
at random, in the plane of reference, 
and let AD be drawn from the point A 
perpendicular to BC. [1. 12] 

If then AD is also perpendicular to 
the plane of reference, that which was 
enjoined will have been done. 

But, if not, let DE be drawn from the point D at right 
angles to BC and in the plane of reference, [1. 1 1] 

let AFbe drawn from A perpendicular to DE, [1. 11] 

and let GH be drawn through the point F parallel to BC. 

['• 3i] 

Now, since BC is at right angles to each of the straight 
lines DA, DE, 

therefore BC is also at right angles to the plane through 
ED, DA. [xi. 4 ] 

And GH is parallel to it ; 
but, if two straight lines be parallel, and one of them be at 
right angles to any plane, the remaining one will also be at 
right angles to the same plane ; [ate 8] 

therefore GH is also at right angles to the plane through 
ED, DA. 



XI. Il] 



PROPOSITIONS 10, n 



'93 



Therefore GH is also at right angles to all the straight 
lines which meet it and are in the plane through ED, DA. 

[xi. Def. 3] 

But AF meets it and is in the plane through ED, DA ; 
therefore GH is at right angles to FA, 
so that FA is also at right angles to GH. 

But AF is also at right angles to DE ; 
therefore AF is at right angles to each of the straight lines 
GH, DE. 

But, if a straight line be set up at right angles to two 
straight lines which cut one another, at the point of section, 
it will also be at right angles to the plane through them ; [xi. 4] 
therefore FA is at right angles to the plane through ED, GH. 

But the plane through ED, GH is the plane of reference ; 
therefore AF is at right angles to the plane of reference. 

Therefore from the given elevated point A the straight 
line AF has been drawn perpendicular to the plane of 
reference. 

Q. E. F, 

The text-books differ in the form which they give to this proposition rather 
than in substance. They commonly assume the construction of a plane 
through the point A at right angles to any straight line BC in the given plane 
{the construction being effected in the manner shown at the end of the note 
on xi. 5 above). The advantage of this method is that it enables a 
perpendicular to be drawn from a point in the plane also, by the same 
construction. (Where the letters for the two figures differ, those referring to 
the second figure are put in brackets.) 

ft 

c 





We can Include the construction of the plane through A perpendicular to 
BC, and make the whole into one proposition, thus. 

BC being any straight line in the given plane MN, draw AD perpendicu- 
lar to JC. 

In any plane passing through BC but not through /( draw DE at right 
angles to BC. 

Through DA, DE draw a plane ; this will intersect the given plane MN 
in a straight line, as FD {AD). 

In the plane AG draw AH perpendicular to FG (AD). 

Then AH is the perpendicular required. 



294 BOOK XI [xi. ii, ii 

In the plane MN, through H in the first figure and A in the second, draw 
KL parallel to BC. 

Now, since BC is perpendicular to both DA and DE, BC is perpendicular 
to the plane AG. [xr. 4] 

Therefore KL, being parallel to BC, is also perpendicular to the plane 
AG [xi..8], and therefore to AH which meets it and is in that plane. 

Therefore AH is perpendicular to both FD (AD) and KL at their point 
of intersection. 

Therefore AH is perpendicular to the plane MM 

Thus we have solved the problem in xt, 12 as well as that in XI. 11; and 
this direct method of drawing a perpendicular to a plane from a point in it is 
obviously preferable to Euclid's method by which the construction of a 
perpendicular to a plane from a point withmtt it is assumed, and a line is 
merely drawn from a point in the plane parallel to the perpendicular obtained 



Proposition 12. 

To set up a straight line at right angles to a given plane 
from a given point in it. 

Let the plane of reference be the given plane, 

and A the point in it; 

thus it is required to set up from the point 
A a straight line at right angles to the 
plane of reference. 

Let any elevated point B be conceived, 

from B let BC be drawn perpendicular to 
the plane of reference, [xi. it] 

and through the point A let AD be drawn 
parallel to BC. [t. 31] 

Then, since AD, CB are two parallel straight lines, 
while one of them, BC, is at right angles to the plane of 
reference, 

therefore the remaining one, AD, is also at right angles to 
the plane of reference. [xi. 8] 

Therefore AD has been set up at right angles to the given 
plane from the point A in ,t. 





xi. 1 3] PROPOSITIONS ii— 13 



Proposition 13. 

From the same point two straight lines cannot be set up at 
right angles to the same plane on the same side. 

For, if possible, from the same point A let the two straight 
lines AB, AC be set up at right 
angles to the plane of reference and on 
the same side, 

and let a plane be drawn through BA, 
AC; 

it will then make, as section through >? 
in the plane of reference, a straight line. 

[xi. 3] 
Let it make DAE ; 
therefore the straight lines AB, AC, DAE are in one plane. 

And, since CA is at right angles to the plane of reference, 
it will also make right angles with all the straight lines which 
meet it and are in the plane of reference. [xi. Def. 3] 

But DAE meets it and is in the plane of reference ; 

therefore the angle CAE is right. 

For the same reason 
the angle BAE is also right ; 
therefore the angle CA E is equal to the angle BAE. 

And they are in one plane : 
which is impossible. 
Therefore etc. 

Q. E. D. 

Sim son added words to this as follows : 

" Also, from a point above a plane there can be but one perpendicular to 
that plane ; for, if there could be two, they would be parallel to one another 
[xi. 6], which is absurd." 

Euclid does not give this result, but we have already had it in the note 
above to xi. 4 {ad fin.). 




j 9 6 BOOK XI [xi. 14 



Proposition 14. 

Planes to which the same straight line is at right angles 
will be parallel. 

For let any straight line AB be at right angles to each of 
the planes CD, EF; 

I say that the planes are 
parallel. / 

For, if not, they will meet °~ 
when produced. r 

Let them meet ; {_ 

they will then make, as 

common section, a straight line. [xi. 3] 

Let them make GH ; 
let a point K be taken at random on GH, 
and let AK, BK be joined. 

Now, since AB is at right angles to the plane EF, 

therefore AB is also at right angles to BK which is a straight 
line in the plane EF produced ; [xi. Def. 3] 

therefore the angle ABK is right. 

For the same reason 
the angle BAK is also right. 

Thus, in the triangle ABK, the two angles ABK, BAK 
are equal to two right angles : 

which is impossible. [1. 17] 

Therefore the planes CD, EF will not meet when 
produced ; 

therefore the planes CD, EFare parallel. [xi. Def. 8] 

Therefore planes to which the same straight line is at right 
angles are parallel. 

Q. E. D. 




xi. 15] PROPOSITIONS 14, 15 197 

Proposition 15. 

If (wo straight lines meeting one another be parallel to two 
straight lines meeting one another, not being in the same plane, 
the planes through them are parallel. 

For let the two straight lines AB, BC meeting one another 
be parallel to the two straight lines 
DE, EF meeting one another, not 
being in the same plane ; 
I say that the planes produced 
through AB, BC and DE, EFw\\\ 
not meet one another. 

For let BG be drawn from the 
point B perpendicular to the plane 
through DE, £F[x.i. 11], and let it 
meet the plane at the point G ; 
through G let GH be drawn 
parallel to ED, and GK parallel to EF. [1. 31] 

Now, since BG is at right angles to the plane through 
DE,EF, 

therefore it will also make right angles with all the straight 
lines which meet it and are in the plane through DE, EF. 

[xi. Def. 3] 

But each of the straight lines GH, GK meets it and is in 
the plane through DE, EF; 
therefore each of the angles BGH, BGK is right. 

And, since BA is parallel to GH, [xi. 9] 

therefore the angles GBA, BGH are equal to two right angles. 

[1. 2 9 ] 
But the angle BGH is right ; 
therefore the angle GBA is also right ; 
therefore GB is at right angles to BA. 

For the same reason 
GB is also at right angles to BC. 

Since then the straight line GB is set up at right angles 
to the two straight lines BA, BC which cut one another, 
therefore GB is also at right angles to the plane through 
BA, BC. [xi. 4] 






*o8 BOOK XI [xi. 15 

But planes to which the same straight line is at right 
angles are parallel ; [xi. 14] 

therefore the plane through AB, BC is parallel to the plane 
through DE t EF. 

Therefore, if two straight lines meeting one another be 
parallel to two straight lines meeting one another, not in the 
same plane, the planes through them are parallel. 

Q. E. D. 

This result is arrived at in the American text-books already quoted by 
starting from the relation between a plane and a straight line parallel to it. 
The series of propositions is worth giving. A straight line and a plane being 
parallel if they do not meet however far they may be produced, we have the 
following propositions. 

t. Any plane containing one t and only one, of two parallel straight lines is 
parallel to the other. 

For suppose AB, CD to be parallel and CD to lie in the plane MN. 

Then A B, CD determine a plane intersecting MN in the straight line CD 

Thus, if AB meets MN, it must meet 
it at some point in CD. fy 

But this is impossible, since AB is 
parallel to CD. 

Therefore A B will not meet the plane 
MN, and is therefore parallel to it. 

[This proposition and the proof are in 
Legend re,] 

The following theorems follow as corollaries. 

2. Through a given straight line a plane can he drawn parallel to any other 
given straight line; and, if the lines are not parallel, only one such plane can he 
drawn. 

We have simply to draw through any point on the first line a straight line 
parallel to the second line and then pass a plane through these two intersecting 
lines. This plane is then, by the above proposition, parallel to the second 
given straight line. 

3. Through a given point a plane can he drawn parallel to any two straight 
lines in space ; and, if the tatter are not parallel, only one such plane can be 
drawn. 

Here we draw through the point straight lines parallel respectively to the 
given straight lines and then draw a plane through the lines so drawn. 
Next we have the partial converse of the first proposition above. 

4. If a straight line is parallel to a plane, it is also parallel to the inter- 
section of any plane through it with the given plane. 

Let AB be parallel to the plane MN, and let 
any plane through AB intersect MN\n CD. 

Now AB and CD cannot meet, because, if 
they did, AB would meet the plane MN, 

And AB, CD are in one plane. 

Therefore AB, CD are parallel. 

From this follows as a corollary : 





xi. is] 



PROPOSITION is 



*99 





5. If each of two intersecting straight lines is parallel to a given plane, 
the plane containing them is parallel to the given 
plane. 

Let AB, AC be parallel to the plane 
MN. 

Tben, if the plane ABC were to meet the 
plane MN, the intersection would be parallel 
both to AB and to AC; which is impossible. 

Lastly, we have Euclid's proposition. 

6. If two straight lines forming an angle are respectively parallel to tivo 
other straight lines forming an angle, the plane of 

the first angle is parallel to the plane of the second. 

Let ABC, DEF be the angles formed by 
straight lines parallel to one another respectively. 

Then, since AB is parallel to DE, 
the plane of DEF is parallel to AB [(r) above]. 

Similarly the plane of DEF is parallel to 
BC. 

Hence the plane of DEF is parallel to the 
plane of ABC [(5)]. 

Legend re arrives at the result by yet another method. He first proves 
Eucl. xi. 16 to the effect that, if two parallel planes are cut by a third, the lines 
of intersection are parallel, and then deduces from this that, (/ two parallel 
straight lines are terminated by tivo parallel planes, the straight lines are equal 
in length, 

(The latter inference is obvious because the plane through the parallels 
cuts the parallel planes in parallel lines, which 
therefore, with the given parallel lines, form a 
parallelogram.) 

Legendre is now in a position to prove 
Euclid's proposition Xi. 15. 

If ABC, DEF be the angles, make AB 
equal to DE, and BC equal to EF, and join 
CA, ED, BE, CF, AD. 

Then, as in Eucl. xi. 10, the triangles 
ABC, DEFzie equal in all respects; 
and AD, BE, CFa.m all equal. 

It is now proved that the planes are 
parallel by reduclio ad absurdum from the 
last preceding result. For, if the plane ABC 

is not parallel to the plane DEF, let the plane drawn through B parallel to the 
plane DEF meet CF, AD in H, G respectively. 

Then, by the last result BE, HF, GD will all be equal. 

But BE, CF, AD are all equal 1 
which is impossible. 

Therefore etc. 




BOOK XI [xi. 16 

Proposition 16. 

If two parallel planes be cut by any plane, their common 
sections are parallel. 

For let the two parallel planes AB, CD be cut by the 
plane EFGH, 

and let EF, GH be their common sections ; 
I say that EF is parallel to GH. 

-)B 







For, if not, EF, GH will, when produced, meet either in 
the direction of F, H or of E, G. 

Let them be produced, as in the direction of F, H, and 
let them, first, meet at K. 

Now, since EFK is in the plane AB, 
therefore all the points on EFK are also in the plane AB. 

[XI. ,] 

But K is one of the points on the straight line EFK; 
therefore K is in the plane AB. 

For the same reason 
K is also in the plane CD ; 
therefore the planes AB, CD will meet when produced. 

But they do not meet, because they are, by hypothesis, 
parallel ; 

therefore the straight lines EF, GH will not meet when 
produced in the direction of F t H. 

Similarly we can prove that neither will the straight lines 
EF, GH meet when produced in the direction of E, G. 

But straight lines which do not meet in either direction 
are parallel. [i. Def. 23] 

Therefore EF is parallel to GH, 

Therefore etc. q. e. d. 



xi. i6, 1 7 J 



PROPOSITIONS 16, 17 



301 



Simson points oat that, in here quoting 1. Def. 23, Euclid should have 
said "But straight lines in one plant which do not meet in either direction are 
parallel" 

From this proposition is deduced the converse of xi. 14. 

If a straight line is perpendicular to one of two parallel planes, it is 
perpendicular to the other also. 

For suppose that MN, PQ are two parallel planus, and that AB is perpen- 
dicular to MN. 

Through AB draw any plane, and let it intersect 
the planes MN, PQ in AC, BD respectively. / -^T~7N 

Therefore AC, BD are parallel. [xi. 16] 

But A C is perpendicular to AB ; Hf" 

therefore AB is also perpendicular to BD. 

That is, AB is perpendicular to any line in PQ 
passing through B ; 
therefore AB is perpendicular to PQ. 

It follows as a corollary that 

Through a given point one plane, and only one, aw be drawn parallel to a 
given plane. 

In the above figure let A be the given point and PQ the given plane. 

Draw AB perpendicular to PQ. 

Through A draw a plane MN at right angles to AB (see note on xi. 5 
above). 

Then MNh parallel to PQ. [xi, 14] 

If there could pass through A a second plane parallel to PQ, AB would 
also be perpendicular to it. 

That is, AB would be perpendicular to two different planes through A : 
which is impossible (see the same note). 

Also it is readily proved that, 

If two planes are parallel to a third plane, they are parallel to one another. 




Proposition 17. 






If two straight lines be cut by parallel planes, (key will be 
cut in the same ratios. 

For let the two straight 
lines AB, CD be cut by the 
parallel planes GH, KL, MN 
at the points A, E, B and C, 
F,D; 

I say that, as the straight line 
AE is to EB, so is CF to FD. 
For let AC, BD, AD be 
joined, 

let AD meet the plane KL 
at the point O, 
and let EO, OF be joined. 




3 oi BOOK XI [xi. 17, 18 

Now, since the two parallel planes KL, MN are cut by 
the plane EBDO, 
their common sections EO, BD are parallel. [xi. 16] 

For the same reason, since the two parallel planes GH, 
KL are cut by the plane AOFC, 
their common sections AC, OF are parallel. [id.] 

* 

And, since the straight line EO has been drawn parallel to 
BD, one of the sides of the triangle ABD, 

therefore, proportionally, as AE is to EB, so is AO to OD. 

[vi. a] 

Again, since the straight line OF has been drawn parallel 
to AC, one of the sides of the triangle ADC, 
proportionally, as AO is to OD, so is CF to FD. [id.] 

But it was also proved that, as AO is to OD, so is AE 
to EB; 
therefore also, as AE is to EB, so is CF to FD. [v. 1 1] 

Therefore etc. 

Q. E. D. 

Proposition 18. 

If a straight line be at right angles to any plane, all the 
planes through it will also be at right angles to the same plane. 

For let any straight line AB be at right angles to the 
plane of reference; 

I say that all the planes through n n m 
AB are also at right angles to the 
plane of reference. 

For let the plane DE be drawn 
through AB, 

let CE be the common section of 
the plane DE and the plane of 
reference, 

let a point F'he. taken at random on CM, 
and from F let FG be drawn in the plane DE at right 
angles to CE. [1. n] 

Now, since AB is at right angles to the plane of reference, 




xi. 18J PROPOSITIONS 17, 18 303 

AB is also at right angles to all the straight lines which meet 
it and are in the plane of reference ; [xi. Def. 3] 

so that it is also at right angles to CE ; 
therefore the angle ABF is right. 

But the angle GFB is also right ; 
therefore AB is parallel to FG. [1. *8] 

But AB is at right angles to the plane of reference ; 
therefore FG is also at right angles to the plane of reference. 

[XL 8] 

Now a plane is at right angles to a plane, when the 
straight lines drawn, in one of the planes, at right angles to 
the common section of the planes are at right angles to the 
remaining plane. [xi. Def. 4] 

And FG, drawn in one of the planes DE at right angles 
to CE, the common section of the planes, was proved to be 
at right angles to the plane of reference ; 

therefore the plane DE is at right angles to the plane of 
reference. 

Similarly also it can be proved that all the planes through 
AB are at right angles to the plane of reference. 
Therefore etc. 

Q. E. D. 

Starting as Euclid does from the definition of perpendicular planes as 
planes such that all straight lines drawn in one of the planes at right angles to 
the common section are at right angles to the other plane, it is necessary for 
him to show that, if F be any point in CE, and FG be drawn in the plane 
DE at right angles to CE, FG will be perpendicular to the plane to which 
AB Is perpendicular. 

It is perhaps more scientific to make the definition, as Legendre makes it, 
a particular case of the definition of the inclination of plants. Perpendicular 
plants would thus be planes such that the angle which (when it is acute) 
Euclid calls the inclination of a plane to a plane is a right angle. When to this 
is added the fact incidentally proved in xi, 10 that the " inclination of a plane to 
a plane " is the same at whatever point in their common section it is drawn, it 
is sufficient to prove the perpendicularity of two planes if one straight line 
drawn, in one of them, perpendicular to their common section is perpendicular 
to the other. 

If this point of view is taken, Props. 18, 19 are much simplified (cf. 
Legendre, H. M. Taylor, Smith and Bryant, Rausenberger, Schultze and 
Seven oak, Holgate), The alternative proof is as follows. 

Let AB be perpendicular to the plane MN, and CE any plane through 
AB, meeting the plane MN'm the straight line CD. 

In the plane MN&sm BF&t right angles to CD. 



304 



BOOK XI 



[xi. 18, 19 



Then ABF'm the angle which Euclid calls (in the case where it is acute) 
the " inclination of the plane to the plane." 







But, since AB is perpendicular to the plane MN. it is perpendicular to 
BF in it. 

Therefore the angle ABF is a right angle ; 
whence the plane CE is perpendicular to the plane MN. 



Proposition 19. 

If two planes which cut one another be at right angles to 
any plane, their common section will also be at right angles to 
the same plane. 

For let the two planes AB, BC be at right angles to the 
plane of reference, 

and let BD be their common section ; 
I say that BD is at right angles to the 
plane of reference. 

For suppose it is not, and from the 
point D let DE be drawn in the plane 
AB at right angles to the straight line 
AD, and DF in the plane BC at right 
angles to CD. 

Now, since the plane AB is at right 
angles to the plane of reference, 

and DE has been drawn in the plane AB at right angles to 
AD, their common section, 

therefore DE is at right angles to the plane of reference. 

[xi. Def. 4] 




xi. 19] PROPOSITIONS 18, 19 305 

Similarly we can prove that 
DF is also at right angles to the plane of reference. 

Therefore from the same point D two straight lines have 
been set up at right angles to the plane of reference on the 
same side : 

which is impossible. [xi. 13] 

Therefore no straight line except the common section DB 
of the planes AB, BC can be set up from the point D at right 
angles to the plane of reference. 

Therefore etc. 

CJ. K. U. 

Legend re, followed by other writers already quoted, uses a preliminary 
proposition equivalent to Euclid's definition of planes at right angles to one 
another. 

// two planes are perpendicular to one another, a straight line drawn in one 
of them perpendicular to their common section -will be perpendicular to the other. 

Let the perpendicular planes CE, MN (figure of last note) intersect in 
CD, and let AB be drawn in CE perpendicular to CD. 

In the plane MN draw BFsX right angles to CD. 

'then, since the planes are perpendicular, the angle A BF (their inclination) 
is a right angle. 

Therefore AB is perpendicular to both CD and BF, and therefore to the 
plane MN. 

We are now in a position to prove XI. 19, viz. If two planes lie perpendicular 
to a third, their intersection is also perpen- 
dicular to that third plane. A^ 

Let each of the two planes AC, AD 
intersecting in A B be perpendicular to the 
plane MN. 

Let A C, AD intersect MN in BC, BD 
respectively. 

In the plane MN draw BE at right 
angles to BC and BF at right angles to 
BD. 

Now, since the planes AC, MN axe at 
right angles, and BE is drawn in the latter perpendicular to BC, BE is 
perpendicular to the plane AC. 

Hence AB is perpendicular to BE. [xi. 4] 

Similarly AB is perpendicular to BF. 

Therefore AB is perpendicular to the plane through BF, BF, i.e. 10 the 
plane MN. 

An useful problem is that of drawing a common perpendicular to two 
straight lines not in one plane, and in connexion with this the following 
proposition may be given. 




3 o6 



BOOK XI 



[xi. 19 



M 



Given a plant and a straight lint not perpendicular to it, one plant, and only 
one, can be drawn through the straight line perpen- 
dicular to the plane. 

Let AB be the given straight line, MN the 
given plane. 

From any point C in AB draw CD perpen- 
dicular to the plane MN. 

Through AB and CD draw a plane AE. 

Then the plane AE is perpendicular to the 
plane MN [xi. 18] 

If any other plane could be drawn through 
AB perpendicular to MN, the intersection AB of 
the two planes perpendicular to MN would itself 
be perpendicular to MN: 
which contradicts the hypothesis. 

To draw a common perpendicular to two straight lines not in the same plane. 

Let AB, CD he the given straight lines. 

Through CD draw the plane MN parallel to AB (Prop, a in note 
to Xi. 15). 

Through AB draw the plane AE perpendicular to the plane MN($ee the 
last preceding proposition). 

AH KB 



[XI. 19] 




Let the planes AE, MN intersect in EF, and let EFm&X CD in G. 

From G, in the plane AE, draw GHai, right angles to EF, meeting AB in H. 

GH is then the required perpendicular. 

For AB it parallel to EF (Prop. 4 in note to xi. 15) ; therefore GH, 
being perpendicular to EF, is also perpendicular to AB, 

But, the plane AE being perpendicular to 'he plane MN, and GH being 
perpendicular to EF, their intersection, 

GH is perpendicular to the plane MN, and therefore to CD. 

Therefore GH is perpendicular to both AB and CD. 

Only one common perpendicular can be drawn to two straight lines not in 
one plane. 

For, if possible, let KL also be perpendicular to both AB and CD. 

Let the plane through KL, AB meet the plane MN in LQ, 

Then AB is parallel to LQ (Prop. 4 in note to xi. 15), so that KL, being 
perpendicular to AB, is also perpendicular to LQ. 

Therefore KL is perpendicular to both CL and LQ, and consequently to 
the plane MN. 

But, if KP be drawn in the plane AF perpendicular to EF, KP is also 
perpendicular to the plane MN. 



XI. 19, 2o] 



PROPOSITIONS 19, 20 



3»7 



Thus there are two perpendiculars from the point K to the plane MN: 
which is impossible. 



Rausenberger's construction for the same problem is more elegant 
he says, through each straight tine a plane parallel 
to the other. Then draw through each straight line 
a plane perpendicular to the plane through the 
other. The two planes last drawn will intersect 
in a straight line, and this straight line is the 
common perpendicular required. 



Draw, 



\ 



\ 



CDF, 




The form of the construction best suited for 
examination purposes, because the most self- 
contained, is doubtless that given by Smith and 
Bryant. 

Let AB, CD be the two given straight lines. 

Through any point E in CD draw EF parallel to AB. 

From any point G in AB draw GH perpendicular to the plane 
meeting the plane in H. 

Through H in the plane CDF draw 
HK parallel to FE or AB, to cut CD 
in K. 

Then, since AB, HK are parallel, 
AGHK is a plane. 

Complete the parallelogram GHKL. 

Now, since LK, GHare parallel, and 
GH is perpendicular to the plane CDF, 

LK is perpendicular to the plane 
CDF. 

Therefore LK is perpendicular to CD and KH, and therefore loAB which 
is parallel to KH 

Proposition 20. 

If a solid angle be contained by three plane angles, any two, 
taken together in any manner, are greater than the remaining 

one. 

For let the solid angle at A be contained by the three 
plane angles BAC, CAD, DAB ; 
I say that any two of the angles 
BAC, CAD, DAB, taken to- 
gether in any manner, are greater 
than the remaining one. 

If now the angles BAC, CAD, 
DAB are equal to one another, 
it is manifest that any two are greater than the remaining one. 

But, if not, let BAC be greater, 
and on the straight line AB, and at the point A on it, let the 




3 o8 BOOK XI [xx jo 

angle BAE be constructed, in the plane through BA, AC, 

equal to the angle DAB ; 

let AE be made equal to AD, 

and let BEC, drawn across through the point E, cut the 

straight lines AB, AC at the points B, C\ 

let DB, DC be joined. 

Now, since DA is equal to AE, 
and AB is common, 
two sides are equal to two sides ; 
and the angle DAB is equal to the angle BAE ; 
therefore the base DB is equal to the base BE. [i. 4] 

And, since the two sides BD, DC are greater than BC, 

[1. »] 
and of these DB was proved equal to BE, 

therefore the remainder DC is greater than the remainder EC, 

Now, since DA is equal to AE, 
and AC is common, 

and the base DC is greater than the base EC, 
therefore the angle DAC is greater than the angle EAC. 

Li- »«] 

But the angle DAB was also proved equal to the angle 
BAE; 

therefore the angles DAB, DAC are greater than the angle 
BAC. 

Similarly we can prove that the remaining angles also, 
taken together two and two, are greater than the remaining 
one. 

Therefore etc 

Q. E. D, 

After excluding the obvious case in which all three angles are equal, 
Euclid goes on to say "If not, let the angle BAC be greater," without adding 
greater than what. Heiberg is clearly right in saying that he means greater 
than BAD, i.e. greater than one of the adjacent angles. This is proved by 
the words at the end " Similarly we can prove," etc. Euclid thus excludes 
as obvious the case where one of the three angles is not greater than either of 
the other two, but proves the remaining cases. This is scientific, but he might 
further have excluded as obvious the case in which one angle is greater than 
one of the others but equal to or less than the remaining one. 



xi. io, si] PROPOSITIONS io, 21 309 

Si m 50 n remarks that the angle BAC may happen to be equal to one of 
the other two and writes accordingly " If they [all three angles] are not [equal], 
let BAC be that angle which is not less than either of the other two, and is 
greater than one of them DAB''' He then proves, in the same way as Euclid 
does, that the angles DAB, DAC are greater than the angle BAC, adding 
finally : "But BAC is not less than either of the angles DAB, DAC; there- 
fore BAC, with either of them, is greater than the other." 

It would he better, as indicated by Legend re and Rausenberger, to begin 
by saying that, "If one of the three angles is either equal to or less than either 
of the other two, it is evident that the sum of those two 15 greater than the 
6rst. It is therefore only necessary to prove, for the case in which one angle is 
greater than each of the others, that the sum of the two latter is greater than 
the former. 

Accordingly let BA C be greater than each of the other angles." We then 
proceed as in Euclid. 

Proposition 21, 

Any solid angle is contained by plane angles less than four 
right angles. 

Let the angle at A be a solid angle contained by the plane 
angles BAC, CAD, DAB ; 
I say that the angles BAC, CAD, 
DAB are less than four right angles. 

For let points B, C, D be taken 
at random on the straight lines AB, 
AC, AD respectively. 
and let BC, CD, DB be joined. B 

Now, since the solid angle at B is contained by the three 
plane angles CBA, ABD, CBD, 

any two are greater than the remaining one ; [xi. 10] 

therefore the angles CBA, ABD are greater than the angle 
CBD. 

For the same reason 
the angles BCA, ACD are also greater than the angle BCD, 
and the angles CD A, ADB are greater than the angle CDS ; 
therefore the six angles CBA, ABD, BCA, ACD, CD A, 
ADB are greater than the three angles CBD, BCD, CDB. 

But the three angles CBD, BDC, BCD are equal to two 
right angles ; [1. 3*] 

therefore the six angles CBA, ABD, BCA, ACD, CD A, 
ADB are greater than two right angles. 




3 io BOOK XI [xi. 21 

And, since the three angles of each of the triangles ABC, 
A CD, ADB are equal to two right angles, 
therefore the nine angles of the three triangles, the angles 
CBA, ACB, BAC, ACD, CD A, CAD, ADB, DBA, BAD 
are equal to six right angles ; 

and of them the six angles ABC, BCA t ACD, CD A, ADB, 
DBA are greater than two right angles ; 
therefore the remaining three angles BAC, CAD, DAB 
containing the solid angle are less than four right angles. 

Therefore etc. 

Q. E. D. 

It will be observed that, although Euclid enunciates this proposition for 
any solid angle, he only proves it for the particular case of a trihedral angle. 
This is in accordance with his manner of proving one case and leaving the 
others to the reader. The omission of the convex polyhedral angle here 
corresponds to the omission, after i. 32, of the proposition about the interior 
angles of a convex polygon given by Prod us and in most books. The proof 
of the present proposition for any convex polyhedral angle can of course be 
arranged so as not to assume the proposition that the interior angles of a 
convex polygon together with four right angles are equal to twice as many 
right angles as the figure has sides. 

Let there be any convex polyhedral angle with V as vertex, and let it be 
cut by any plane meeting its faces in, say, the 
polygon ABCDE. 

Take O any point within the polygon, and 
in its plane, and join OA, OB, OC, OD, OE. 

Then all the angles of the triangles with 
vertex O are equal to twice as many right angles 
as the polygon has sides ; [1. 3a] 

therefore the interior angles of the polygon to- 
gether with all the angles round are equal to 
twice as many right angles as the polygon has 
sides. 

Also the sum of the angles of the triangles 
VAB, VBC, etc, with vertex P'are equal to twice as many right angles as the 
polygon has sides ; 

and all the said angles are equal to the sum of (1) the plane angles at V 
forming the polyhedral angle and (2) the base angles of the triangles with 
vertex V. 

This latter sum is therefore equal to the sum of (3) all the angles 
round O and (4) all the interior angles of the polygon. 

Now, by Euclid's proposition, of the three angles forming the solid angle at 
A, the angles VAE, VAB are together greater than the angle EAB. 

Similarly, at B, the angles VBA, VBC are together greater than the angle 
ABC. 

And so on. 

Therefore, by addition, the base angles of the triangles with vertex V 




xi. it] PROPOSITION zi 311 

!(x) above] are together greater than the sum of the angles of the polygon 
(4) above]. 

Hence, by way of compensation, the sum of the plane angles at V[(i) 
above] is less than the sum of the angles round [(3) above]. 

But the latter sum is equal to four right angles; therefore the plane angles 
forming the polyhedral angle are together less than four right angles. 

The proposition is only true of convex polyhedral angles, i.e. those in 
which the plane of any face cannot, if produced, ever cut the solid angle. 

There are certain propositions relating to equal (and symmetrical) trihe- 
dral angles which are necessary to the consideration of the polyhedra dealt 
with by Euclid, all of which (as before remarked) have trihedral angles only. 

1. Two trihedral angles are equal if two face angles and the included 
dihedral angle of the one are respectively equal to two face angles anil the included 
dihedral angle of the other, the equal parts being arranged in the same order. 

2. Two trihedral angles are equal if two dihedral angles and the included 
face angle of the one are respectively equal to two dihedral angles and the included 
fate angle of the other, all equal parts being arranged in the same order. 

These propositions are proved immediately by superposition. 

3. Two trihedral angles are equal if the three face angles of the one are 
respectively equal to the three face angles of the other, and all are arranged in the 
same order. 

Let V— ABC and V—A'B'C be two trihedral angles such that the angle 
A VB is equal to the angle A' VB', the angle BVC to the angle B'V'C, and 
the angle CVA to the angle C V'A'. 



We first prove that corresponding pairs of face angles include equal dihedral 
angles. 

E.g, the dihedral angle formed by the plane angles CVA, AVB is equal 
to that formed by the plane angles C V A , A' VB'. 

Take points A, B, C on VA, VB, VC and points A', If, C on V'A', 
VB 1 , VC, such that VA, VB, VC. V'A', VB\ VC are all equal. 

Join BC, CA, AB, SC, C'A\ A'B'. 

Take any point Don AV, and measure A'D' along A' V equal to AD. 

From D draw DE in the plane AVB, and DF in the plane CVA, 
perpendicular to A V. Then DE, DF will meet AB, AC respectively, the 
angles VAB, VAC, the base angles of two isosceles triangles, being i-^ss than 
right angles. 

Join EF. 

Draw the triangle DEF in the same way. 



312 BOOK XI [XI. 21, 22 

Now, by means of the hypothesis and construction, it appears that the 
triangles VAB, V'A'B" are equal in all respects. 

So are the triangles VAC, VA'C', and the triangles VBC, VBC. 

Thus BC, CA, AB are respectively equal to B'C, C'A\ A'B", and the 
triangles ABC, A'ffC are equal in all respects. 

Now, in the triangles ADE, A'DE, 
the angles ADE, DAE are equal to the angles A'DE, DA'S respectively, 
and AD is equal to A'D. 

Therefore the triangles ADE, A'DE are equal in all respects. 

Similarly the triangles ADE, A'DE are equal in all respects. 

Thus, in the triangles AEE, A'E'F', 
EA, AEsre respectively equal to EA', A'E' t 
and the angle EAF is equal to the angle E'A'E (from above) ; 
therefore the triangles AEE, A' E E are equal in all respects. 

Lastly, in the triangles DEE, DE'F, the three sides are respectively 
equal to the three sides * 
therefore the triangles are equal in all respects. 

Therefore the angles EDE, EDE" are equal. 

But these angles are the measures of the dihedral angles formed by the 
planes CVA, A VB and by the planes CVA', A' VB" respectively. 
Therefore these dihedral angles are equal. 

Similarly for the other two dihedral angles. 

Hence the trihedral angles coincide if one is applied to the other ; 
that is, they are equal. 

To understand what is implied by " taken in the same order " we may 
suppose ourselves to be placed at the vertices, and to take the faces in clock- 
wise direction, or the reverse, for both angles. 

If the face angles and dihedral angles are taken in reverse directions, i.e. 
in clockwise direction in one and in counterclockwise direction in the other, 
then, if the other conditions in the above three propositions are fulfilled, the 
trihedral angles are not equal but symtnetrical. 

If the faces of a trihedral angle be produced beyond trie vertex, they form 
another trihedral angle. It is easily seen that these vertical trihedral angles 
are symmetrical. 



Proposition 22. 

Jf there be three plane angles of "which two, taken together 
in any manner, are greater than the remaining one, and they 
are contained by equal straight lines, it is possible to construct 
a triangle out of the straight lines joining the extremities of 
the equal straight lines. 

Let there be three piane angles ABC, DEF, GHK, of 



XI. 2 2] 



PROPOSITIONS a i, 22 



3«3 



which two, taken together in any manner, are greater than 
the remaining one, namely 

the angles ABC, DEF greater than the angle GHK, 
the angles DEF, GHK greater than the angle ABC, 

and, further, the angles GHK, ABC greater than the angle 

DEF; 

let the straight lines AB, BC, DE, EF, GH, HK be equal, 

and let AC, DF, GK be joined ; 

I say that it is possible to construct a triangle out of straight 
lines equal to AC, DF, GK, that is, that any two of the 
straight lines AC, DF, GK are greater than the remaining 
one. 







Now, if trre angles ABC, DEF, GHK are equal to one 
another, it is manifest that, AC, DF, GK being equal also, 
it is possible to construct a triangle out of straight lines equal 
to AC, DF,GK. 

But, if not, let them be unequal, 
and on the straight line HK, and at the point H on it, let 
the angle KHL be constructed equal 
to the angle ABC; 
let HL be made equal to one of the 
straight lines AB, BC, DE, EF, GH, 
HK, 
and let KL, GL be joined. 

Now, since the two sides AB, BC 
are equal to the two sides KH, HL, 
and the angle at B is equal to the angle KHL, 
therefore the base AC is equal to the base KL. 

And, since the angles ABC, GHK are greater than the 
angle DEF, 




[••4] 



3i4 BOOK XI [xi. it, *i 

while the angle ABC is equal to the angle KHL, 
therefore the angle GHL is greater than the angle DEF. 

And, since the two sides GH, HL are equal to the two 
sides DE, EF, 

and the angle GHL is greater than the angle DEF, 
therefore the base GL is greater than the base DF. [t. 84] 

But GK, KL are greater than GL. 

Therefore GK, KL are much greater than DF. 

But KL is equal to AC; 
therefore AC. GK are greater than the remaining straight 
line DF. 

Similarly we can prove that 
AC, DFare greater than GK, 
and further DF, GK are greater than A C. 

Therefore it is possible to construct a triangle out of 
straight lines equal to AC, DF, GK. 

q. e. d, 

The Greek text gives an alternative proof, which is relegated by Heiberg 
to the Appendix. Simson selected the alternative proof in preference to that 
given above ; he objected however to words near the beginning, " If not, let 
the angles at the points B, E, H be unequal and that at B greater than either 
of the angles at B, H," and altered the words so as to take account of the 
possibility that the angle at B might be equal to one of the other two. 

As will be seen, Euclid takes no account of the relative magnitude of the 
angles except as regards the case when all three are equal. Having proved 
that one base is less than the sum of the two others, he says that " similarly 
we can prove" the same thing for the other two bases. 

If a distinction is to be made according to the relative magnitude of the 
three angies, we may say, as in the corresponding place in xi. 21, that, if one 
of the three angles is either equal to or less than either of the other two, the 
bases subtending those two angles must obviously be together greater than the 
base subtending the first. Thus it is only necessary to prove, for the case in 
which one angle is greater than either of the others, that the sum of the bases 
subtending those others is greater than that subtending the first. This is 
practically the course taken in the interpolated alternative proof. 

Proposition 23. 

To construct a solid angle out of three plane angles two oj 
which, taken together in any manner, are greater than the 
remaining one : thus the three angles must be less than four 
right angles. 



» *$] 



PROPOSITIONS 22, 23 



315 



Let the angles ABC, DEF, GHK be the three given 
plane angles, and let two of these, taken together in any 
manner, be greater than the remaining one, while, further, 
the three are less than four right angles ; 

thus it is required to construct a solid angle out of angles 
equal to the angles ABC, DBF, GHK. 





Let AB, BC, DE, EF, GH, HK be cut off equal to one 
another, 

and let AC, DF, GK be joined ; 

it is therefore possible to construct a triangle out of straight 
lines equal to AC, DF, GK. [xi. is] 

Let LMN be so constructed that 
AC is equal to LM, DF to MN,an<\ 
further GK to NL, 
let the circle LMNbe described about 
the triangle LMN, 
let its centre be taken, and let it be 0; 
let LO, MO, NO be joined ; 
I say that AB is greater than LO. 




For, if not, AB is either equal to LO, or less. 

First, let it be equal. 

Then, since AB is equal to LO, 

while AB is equal to BC, and OL to OM, 

the two sides AB, BC are equal to the two sides LO. OM 
respectively ; 

and, by hypothesis, the base AC is equal to the base LM ; 
therefore the angle ABC is equal to the angle L OM. [1. 8] 

For the same reason 
the angle DEF is also equal to the angle MON, 
and further the angle GHK to the angle NOL ; 



3»6 BOOK XI [xi. 23 

therefore the three angles ABC, DEF, GHK are equal to 
the three angles LOM, MON, NOL, 

But the three angles LOM, MON, NOL are equal to 
four right angles ; 

therefore the angles ABC, DEF, GHK are equal to four 
right angles. 

But they are also, by hypothesis, less than four right angles : 
which is absurd. 

Therefore AB is not equal to LO. 

I say next that neither is AB less than LO. 

For, if possible, let it be so, 
and let OP be made equal to AB, and OQ equal to BC, 
and let PQ be joined. 

Then, since AB is equal to BC, 
OP is also equal to OQ, 
so that the remainder LP is equal to QM. 

Therefore LM is parallel to PQ, (vi. a] 

and LMO is equiangular with PQO ; [1. 39] 

therefore, as OL is to LM, so is OP to PQ ; [vi. 4] 

and alternately, as LO is to OP, so is LM to PQ. [v. 16] 

But LO is greater than OP ; 
therefore LM is also greater than PQ. 

But LM was made equal to AC; 
therefore AC is also greater than PQ. 

Since, then, the two sides AB, BC are equal to the two 

sides PO, OQ, 

and the base AC is greater than the base PQ, 

therefore the angle ABC is greater than the angle POQ. 

['■ 2 S] 
Similarly we can prove that 

the angle DEF is also greater than the angle MON, 

and the angle GHK greater than the angle NOL. 

Therefore the three angles A BC, DEF, GHK ak greater 
than the three angles LOM, MON, NOL. 

But, by hypothesis, the angles ABC, DEF, GHK are 
less than four right angles; 

therefore the angles LOM, MON, NOL are much less than 
four right angles. 



xi. 23] PROPOSITION 23 317 

But they are also equal to four right angles : 

which is absurd. 

Therefore AB is not less than LO. 

And it was proved that neither is it equal ; 

therefore AB is greater than LO. 

Let then OR be set up from the point at right angles 
to the plane of the circle LMN, [xi. u] 

and let the square on OR be equal to that area by which 
the square on AB is greater than the square on LO ; [Lemma] 
let RL, RM, RN be joined. 

Then, since RO is at right angles to the plane of the circle 

LMN, 

therefore RO is also at right angles to each of the straight 
lines LO, MO, NO. 

And, since LO is equal to OM, 

while OR is common and at right angles, 

therefore the base RL is equal to the base RM. [1. 4] 

For the same reason 

RN is also equal to each of the straight lines RL, RM ; 

therefore the three straight lines RL, RM, RN are equal to 
one another. 

Next, since by hypothesis the square on OR is equal to 
that area by which the square on AB is greater than the 
square on LO, 
therefore the square on A B is equal to the squares on LO, OR, 

But the square on LR is equal to the squares on LO, OR, 
for the angle LOR is right ; [1. 47] 

therefore the square on AB is equal to the square on RL ; 
therefore AB is equal to RL. 

But each of the straight lines BC, DE, EF, GH, HK'is 
equal to AB, 

while each of the straight lines RM, RN is equal to RL ; 
therefore each of the straight lines AB, BC, DE, EF, GH, 
HK is equal to each of the straight lines RL, RM, RN. 



3 i8 BOOK XI [xi. *3 

And, since the two sides LR, RM are equal to the two 
sides AB, BC, 

and the base LM is by hypothesis equal to the base A C, 
therefore the angle LRM is equal to the angle ABC. [t- 8] 

For the same reason 
the angle MRN is also equal to the angle DEF, 
and the angle LRN to the angle GHK. 

Therefore, out of the three plane angles LRM, MRN, 
LRN, which are equal to the three given angles ABC, DEF, 
GHK, the solid angle at R has been constructed, which is 
contained by the angles LRM, MRN, LRN. 

Q, E. F. 

Lemma. 

But how it is possible to take the square on OR equal to 
that area by which the square on AB is 
greater than the square on LO, we can show 
as follows. 

Let the straight lines AB, LO be 
set out, 
and let AB be the greater ; 
let the semicircle ABC be described on AB, 
and into the semicircle ABC let AC be fitted equal to the 
straight line LO, not being greater than the diameter AB;\yv. i] 
let CB be joined 

Since then the angle ACB is an angle in the semicircle 
ACB, 
therefore the angle ACB is right. [hi. 31] 

Therefore the square on AB is equal to the squares on 
AC,CB. [1.47] 

Hence the square on AB is greater than the square on 
AC by the square on CB. 

But AC is equal to LO. 

Therefore the square on AB is greater than the square on 
LO by the square on CB. 

If then we cut off OR equal to BC, the square on AB will 
be greater than the square on LO by the square on OR. 

Q. E. F. 

The whole difficulty in this proposition is the proof of a fact which makes 
the construction possible, viz. the fact that, if LMN\& a triangle with sides 




xi. a 3 ] PROPOSITION a 3 319 

respectively equal to the bases of the isosceles triangles which have the 
given angles as vertical angles and the equal sides all of the same length, then 
one of these equal sides, as AB, is greater than the radius LO of the circle 
circuit) scribing the triangle LMN 

Assuming that AB is greater than LO, we have only to draw from O a 
perpendicular OR to the plane of the triangle LMN, to make OR of such a 
length that the sum of the squares on LO, OR is equal to the square on AB, 
and to join RL, RM, RN. (The manner of finding OR such that the square 
on it is equal to the difference between the squares on AB and LO is shown 
in the Lemma at the end of the text of the proposition. We have already 
had the same construction in the Lemma after x. 1 3.) 

Then clearly RL, RM, RN are equal to AB and to one another [1. 4 
and 1. 47]. 

Therefore the triangles LRM, MRN, NRL have their three sides 
respectively equal to those of the triangles ABC, DEF, GHK respectively. 

Hence their vertical angles are equal to the three given angles respectively; 
and the required solid angle is constructed. 

We return now to the proposition to be proved as a preliminary to the 
construction, viz. that, in the figures, AB is greater than LO. 

It will be observed that Euclid, as his manner is, proves it for one case 
only, that, namely, in which 0, the centre of the circle circumscribing the 
triangle LMN, falls within the triangle, leaving the other cases for the reader 
to prove. As usual, however, the two other cases ate found in the Greek text, 
after the formal conclusion of the proposition, as above, ending with the words 
Swtp (Set iroi>/<rai. This position for the proofs itself suggests that they are not 
Euclid's but are interpolated ; and this is rendered certain by the fact that 
words distinguishing three cases at the point where the centre of the 
circumscribing circle is found, " It [the centre] will then be either within the 
triangle LMN or on one of its sides or without. First let it be within," are 
found in the mss. B and V only and are manifestly interpolated. Nevertheless 
the additional two cases must have been inserted very early, as they are found 
in all the best mss. 

In order to give a clear view ol the proof of all three cases as given in the 
text, we will reproduce all three (Euclid's as well as the others) with abbrevia- 
tions to make them catch the eye better. 

In all three cases the proof is by reduetio ad cbsurdum, and it is proved 
first that AB cannot be equal to LO, and secondly that AB cannot be less 
than LO. 

Case I. 

(1) Suppose, if possible, that AB - LO, 

Then AB, BC are respectively equal to LO, OM; 
and AC - LM(by construction). 

Therefore 4 ABC - u LOM. 

Similarly L DEF= l MON, 

i-GHK=t.NOL, 
Adding, we have 

lABC + l DEF+ L GHK= U LOM+ L MON + L NOL 
m four right angles : 
which contradicts the hypothesis. 
Therefore AB*LO. 



320 



BOOK XI 



[xi. 23 



(a) Suppose that AS < LO. 

Make OP, OQ (measured along OL, OM) each equal to AB. 

Thus, OL, OM being equal also, it follows that 
PQ is || to LM. 

Hence LM ': PQ = LO i OP; 

and, since LO > OP, 

LM,i,e.AC,> PQ. 

Thus, in As POQ, ABC, two sides aie equal to two sides, and base 
AC> base PQ; 

therefore l ABC > l POQ, i.e. L LOM. 

Similarly l. DEF> l. MON, 

L GHK> L NOL, 
and it follows by addition that 

l ABC+l. DEF+ l GHK> (four right angles) • 
which again contradicts the hypothesis. 



Case II. 

(i) Suppose, if possible, that AB~ LO. 

Then (AB + BC), or (DE + EF) = M0+0L 

= MN 
= DF: 
which contradicts the hypothesis. 
(2) The supposition that AB < LO is even more 
impossible ; for in this case it would result that 
DE + EF<. DF. 




Case III. 

(1) Suppose, if possihle, that AB = LO. 

Then, in the triangles ABC, LOM, two sides AB, BC are respectively 
equal to two sides LO, OM, and the bases 
A C, LM are equal ; 
therefore l ABC= t- LOM. 

Similarly lGHK=lNOL. 
Therefore, by addition, 

L MON= L ABC+ l GHK 

>i.DEF(by hypothesis). 
But, in the triangles DEF, MON, which 
are equal in all respects, 

i. M0N= l DEF 
But it was proved that L MON> L DEF: 
which is impossible. 

(2) Suppose, if possible, that AB < LO. 
Along OL, OM measure OP, OQ each equal to AB, 




XI. »J] 



PROPOSITION 23 



3 al 






Then LM, PQ are parallel, and 

LM:PQ=LO: OP, 
whence, since LO> OP, 

LM, or AC, > .P(?. 
Thus, in the triangles ABC, POQ, 

lA£C>lPOQ, i.e. lLOM. 
Similarly, by taking OR along ON equal to .4.5, we prove that 

i-GHK>t-LON. 
Now, at 0, make lPOS equal to lABC, and lPOT equal to 

Make OS, CTeach equal to 0/", and join ST, SP, TP. 
Then, in the equal triangles ABC, POS, 

AC=PS, 

so that LM= PS. 

Similarly LN - PT. 

Therefore in the triangles MLN, SPT, since lMLN> lSPT [this is 
assumed, but should have been explained], 

MAT> ST, 
or DF>ST. 

Lastly, in As DEF, SOT, which have two sides equal to two sides, since 
DF> ST, 

l L>EF> l SOT 

> L ABC + l GHK (by construction) : 
which contradicts the hypothesis. 

Simson gives rather different proofs for all tliree cases ; but the essence of 
them can be put, I think, a little more shortly than in his text, as well as more 
clearly 

Case I. (O within ALMN.) 
(i) Let AB be, if possible, equal to LO. 

Then the as ABC, DEF, GHK must be identically equal to the As 
LOM, AfON, NOL respectively. 

H 








K ^6^ F 

Therefore the vertical angles at in the 

latter triangles are equal respectively to the angles 

at B, E, H. 

The tatter are therefore together equal to four 

right angles ; 

which is impossible. 

(2) If AB be less than LO, construct on the 

bases LM, MN, NL triangles with vertices 

P, Q, R and identically equal to the As ABC, 

DEF, GHK respectively. 




3« BOOK XI [xi. 13 

Then P, Q, R will fall within the respective angles at O, since PL = PM 
and < LO, and similarly in the other cases. 

Thus [1. ii] the angles at P, Q, R are respectively greater than the angles 
at O in which they lie. 

Therefore the sum of the angles at P, Q, R, i.e. the sum of the angles at 
B, E, H, is greater than four right angles : 
which again contradicts the hypothesis. 

Case II. {O lying on MN.) 

In this case, whether (1) AS - LO, or (2) A B < LO, a triangle cannot 
be formed with MN as base and each of the other sides equal to AB. In other 
words, the triangle DEF either reduces to a straight line or is impossible. 

H 



Case III. (O lying outside the &LMN.) 

(1) Suppose, if possible, that AB = LO. 

Then the triangles LOM, MON, NOL are identically equal to the 
triangles ABC, DEF, GHK. 

Since L LOM+ 1. LON= l MON, 

lABC + lGHK=lDEF: 

which contradicts the hypothesis. 

(a) Suppose that AB < OL. 

Draw, as before, on LM, MN, NL as bases triangles with vertices P, Q, R 
and identically equal to the As ABC, DEF, GHK. 

Next, at N on the straight line NR, make L RNS equal to the angle 
PLM, cut off NS equal to Z^and join RS, LS. 

Then A NRSis identically equal to A LPAfot A ABC, 

Now ( l LNR + L RNS) < ( i, NLO + l. OLM), 

that is, i. LNS < l NLM. 

Thus, in As LNS, NLM, two sides are equal to two sides, and the included 
angle in the former is less than the included angle in the other. 

Therefore LS<MN. 



xi. 13, 34] 



PROPOSITIONS 33^ 34 



3*3 



Hence, in the triangles MQN, LRS, two sides are equal to two sides, and 
MN>LS.. 



Therefore 



That is, 
which is impossible. 



l MQN ■> l. LRS 

>(lLRN+lSRN) 
> \ l LRN± l. LPM). 

l. DEF> ( l. GHK+ L ABC) : 




Proposition 24. 

If a solid be contained by parallel planes, the opposite planes 
in it are equal and parallelogrammic. 

For let the solid CDHG be contained by the parallel planes 
AC, GF, AH. DF, BF t A£; 
I say that the opposite planes 
in it are equal and parallelo- 
grammic. 

For, since the two parallel 
planes BG, CE are cut by the 
plane AC, 

their common sections are 
parallel. [xl 16] 




314 BOOK XI [xi. 34 

Therefore AB is parallel to DC. 

Again, since the two parallel planes BF, AE are cut by 
the plane AC, 
their common sections are parallel. [xi. 16] 

Therefore BC is parallel to AD. 

But AB was also proved parallel to DC ; 
therefore A C is a parallelogram. 

Similarly we can prove that each of the planes DF, FG, 
GB, BF, AE is a parallelogram. 

Let AH, DF be joined. 

Then, since AB is parallel to DC, and BH to CF, 
the two straight lines AB, BH which meet one another are 
parallel to the two straight lines DC, CF which meet one 
another, not in the same plane ; 

therefore they will contain equal angles ; [xi. 10] 

therefore the angle ABH is equal to the angle DCF. 

And, since the two sides AB, BH are equal to the two 
sides DC, CF, [1. 34] 

and the angle ABH is equal to the angle DCF, 
therefore the base AH 1$ equal to the base DF, 
and the triangle ABH is equal to the triangle DCF, [1. 4] 

And the parallelogram BG is double of the triangle ABH, 
and the parallelogram CE double of the triangle DCF; [1. 34] 
therefore the parallelogram BG is equal to the parallelo- 
gram CE. 

Similarly we can prove that 
A C is also equal to GF, 
and AE to BF. 

Therefore etc. 

Q. E. D. 

As Heiberg says, this proposition is carelessly enunciated. Euclid means 
a solid contained by six planes and not more, the planes are parallel two and 
two, and the opposite faces are equal in the sense of identically equal, or, as 
Simson puts it, equal and similar. The similarity is necessary in order to 
enable the equality of the parallelepipeds in the next proposition to be inferred 
from the roth definition of Book xi. Hence a better enunciation would be: 

If a solid be contained by six planes parallel two and two, the opposite faces 
respectively are equal and similar parallelograms. 

The proof is simple and requires no elucidation. 



Xi. 2$] PROPOSITIONS 24, 25 325 



Proposition' 25. 

If a parallelepipedal solid be cut by a plane "which is 
parallel to the opposite planes, then, as the base is to the base, so 
wilt the solid be to the solid. 

For let the parallelepipedal solid A BCD be cut by the 
plane FG which is parallel to the opposite planes RA, DH \ 

I say that, as the base A EFV is to the base ENCF, so is the 
solid ABFUX.Q the solid EGCD. 




L K A e 

For let AH be produced in each direction, 

let any number of straight lines whatever, AK, KL, be made 
equal to AE, 

and any number whatever, HM, MN, equal to EH; 

and let the parallelograms LP, KV, HW, MS and the solids 
LQ, KR, DM, MTbe completed. 

Then, since the straight lines LK, KA, AE are equal to 
one another, 

the parallelograms LP, KV, A Fare also equal to one another, 
KO, KB, AG are equal to one another, 

and further LX, KQ, AR are equal to one another, for they 
are opposite. [xi. 24] 

For the same reason 

the parallelograms EC, HW, MS are also equal to one another, 

HG, HI, IN are equal to one another, 

and further DH, MY, NT are equal to one another. 

Therefore in the solids LQ, KR, AU three planes are 
equal to three planes. 



3*6 BOOK XI [xi. 25 

But the three planes are equal to the three opposite ; 
therefore the three solids LQ, KR, AU are equal to one 
another. 

For the same reason 
the three solids ED, DM, MT are also equal to one another. 

Therefore, whatever multiple the base LF is of the base 
AF t the same multiple also is the solid L U of the solid A U. 

For the same reason, 
whatever multiple the base NF is of the base FH, the same 
multiple also is the solid NU of the solid HU. 

And, if the base LF is equal to the base NF, the solid L U 
is also equal to the solid NU; 

if the base LF exceeds the base NF, the solid LU also 
exceeds the solid NU; 
and, if one falls short, the other falls short. 

Therefore, there being four magnitudes, the two bases 
AF, FH, and the two solids A U, UN, 
equimultiples have been taken of the base AF and the solid 
A U, namely the base LF and the solid L U, 
and equimultiples of the base NF and the solid NU, namely 
the base NF and the solid NU, 

and it has been proved that, if the base LF exceeds the base 
FN, the solid LU also exceeds the solid NU, 
if the bases are equal, the solids are equal, 
and if the base falls short, the solid falls short. 

Therefore, as the base AF is to the base FN, so is the 
solid AUto the solid UN. [v, Def. 5] 

Q. E D. 

It is to be observed that, as the word p arollelogrammie was used in Book I. 
without any definition of its meaning, so n-opoXX'ifA.tTtiriSos, paralkkpipedal, is 
here used without explanation. While it means simply "with parallel planes," 
i.e. " faces," the term is appropriated to the particular solid which has six 
plane faces parallel two and two. The proper translation of trMpeo* 
TapaWifXtriirtSoir is paralUlepipedal solid, not solid paralltkpiped, as it is 
usually translated. Still less is the solid a parallelepiped, as the word is not 
uncommonly written. 

The opposite faces in each set of parallelepipedal solids in this proposition 
are not only equal but equal and similar. Euclid infers that the solids in each 
set are equal from Def. 10; but, as we have seen in the note on Deft*. 9, io, 



xi. is, z6] 



PROPOSITIONS as. *& 



3*7 



though it is true, where no solid angle in the figures is contained by more 
than three plane angles, that two solid figures are equal and simitar which are 
contained by the same number of equal and similar faces, similarly arranged, 
the fact should have been proved. To do this, we hare only to prove the 
proposition, given above in the note on XI. z I, that two trihedral angles are 
equal if the three face angks of the one are respectively equal to the three face 
angles in the other, and all are arranged in the same order, and then to prove 
equality by applying one figure to the other as is done by Simson in his 
proposition C. 

Application will also, of course, establish what is assumed by Euclid of 
the solids formed by the multiples of the original solids, namely that, if 
LF% NF, the solid ZC/% the solid JVCl. 



Proposition 26. 

On a given straight tine, and at a given point on it, to 
construct a solid angle equal to a given solid angle. 

Let AB be the given straight line, A the given point on 
it, and the angle at D, contained by the angles EDC, EDF, 
FDC, the given solid angle ; 

thus it is required to construct on the straight line AB, and at 
the point A on it, a solid angle equal to the solid angle at D. 







For let a point F be taken at random on DF, 
let FG be drawn from F perpendicular to the plane through 
ED, DC, and let it meet the plane at G, [xi. u] 

let DG be joined, 

let there be constructed on the straight line AB and at the 
point A on it the angle BAL equal to the angle EDC, and 
the angle BAK equal to the angle EDG, [1. 23] 

let AK be made equal to DG, 



338 BOOR XI [xi. s6 

let KH be set up from the point K at right angles to the 
plane through BA, AL, [xi. uj 

let KH be made equal to GF, 

and let HA be joined ; 

I say that the solid angle at A, contained by the angles BAL, 
BAH, HAL is equal to the solid angle at D contained by 
the angles EDC, EDF, FDC. 

For let AB, DE be cut off equal to one another, 
and let HB, KB, FE, GE be joined. 

Then, since FG is at right angles to the plane of reference, 
it will also make right angles with all the straight lines which 
meet it and are in the plane of reference ; [xt Def- 3] 

therefore each of the angles FGD, FGE is right. 

For the same reason 
each of the angles HKA, HKB is also right. 

And, since the two sides KA, AB are equal to the two 
sides GD, DE respectively, 

and they contain equal angles, 

therefore the base KB is equal to the base GE. [1. 4] 

But KH is also equal to GF, 
and they contain right angles ; 
therefore HB is also equal to FE. [1. 4] 

Again, since the two sides AK, KH are equal to the two 
sides DG, GF, 

and they contain right angles, 

therefore the base AH is equal 10 the base FD. [1. 4] 

But AB is also equal to DE ; 
therefore the two sides HA, AB are equal to the two sides 
DF, DE. 

And the base HB is equal to the base FE ; 
therefore the angle BAH is equal tc the angle EDF. [1. 8] 

For the same reason 
the angle HAL is also equal to the angle FDC. 

And the angle BAL is also equal to the angle EDC. 



xi. i6, j 7] PROPOSITIONS 26, rj 339 

Therefore on the straight line AB, and at the point A on 
it, a solid angle has been constructed equal to the given solid 
angle at D. 

Q. E. F. 

This proposition again assumes trie equality of two trihedral angles which 
have the three plane angles of the one respectively equal to the three plane 
angles of the other taken in the same order. 



Proposition 27. 

On a given straight line to describe a parallelepipedal solid 
simitar and similarly situated to a given parallelepipedal solid. 

Let AB be the given straight line and CD the given 
parallelepipedal solid ; 

thus it is required to describe on the given straight line AB 
a parallelepipedal solid similar and similarly situated to the 
given parallelepipedal solid CD, 




M 



For on the straight line AB and at the point A on it let 
the solid angle, contained by the angles BA H, HAK, KAB, 
be constructed equal to the solid angle at C, so that the angle 
BAH is equal to the angle ECF, the angle BAK equal to 
the angle ECG, and the angle KAH to the angle GCF ; 
and let it be contrived that, 
as EC is to CG, so is BA to AK, 
and, as GC is to CF, so is KA to AH, [vi. 1*] 

Therefore also, ex aequali, 
as EC is to CF, so is BA to AH. [v. «] 

Let the parallelogram HB and the solid ALhe completed. 

Now since, as EC is to CG, so is BA to AK, 
and the sides about the equal angles ECG, BAK are thus 
proportional, 



33P BOOK Xt [xi. *7, a8 

therefore the parallelogram GE is similar to the parallelo- 
gram KB. 

For the same reason 
the parallelogram KH is also similar to the parallelogram GF, 
and further FE to HB ; 

therefore three parallelograms of the solid CD are similar to 
three parallelograms of the solid AL. 

But the former three are both equal and similar to the 
three opposite parallelograms, 

and the latter three are both equal and similar to the three 
opposite parallelograms ; 

therefore the whole solid CD is similar to the whole solid AL. 

[xi. Def. 9] 
Therefore on the given straight line AB there has been 
described AL similar and similarly situated to the given 
parallelepipedal solid CD. 

Q. E. F. 



Proposition 28. 

If a parallelepipedal solid be cut by a plane through the 

diagonals of the opposite planes, the solid will be bisected by the 
plane. 

For let the parallelepipedal solid AB be cut by the plane 
CDEF through the diagonals CF, DE of 
opposite planes ; 

I say that the solid AB will be bisected by 
the plane CDEF. 

For, since the triangle CGF is equal 
to the triangle CFB, [1. 34] 

and ADE to DEH, 
while the parallelogram CA is also equal 
to the parallelogram EB, for they are opposite, 
and GE to CH, 

therefore the prism contained by the two triangles CGF, 
ADE and the three parallelograms GE, AC, CE is also equal 
to the prism contained by the two triangles CFB, DEH and 
the three parallelograms CH, BE, CE ; 




xi. *8] PROPOSITIONS if, 28 33« 

for they are contained by planes equal both in multitude and 

in magnitude. [xi. Def. 10] 

Hence the whole solid AB is bisected by the plane CDEF. 

Q. E. D. 

Simson properly observes that it ought to be proved that the diagonals of 
two opposite faces are in one plane, before we speak of drawing a plane 
through them. Clavius supplied the proof, which is of course simple enough. 

Since EF, CD are both parallel to AG or BH, they are parallel to one 
another. 

Consequently a plane can be drawn through CD, EF and the diagonals 
DE, CF are in that plane [xi. 7]. Moreover CD, EF are equal as well as 
parallel ; so that CF, DE are also equal and parallel. 

Simson does not, however, seem to have noticed a more serious difficulty. 
The two prisms are shown by Euclid to be contained by equal faces — the faces 
are in fact equal and similar — and Euclid then infers at once that the prisms 
are equal. But they are not equal in the only sense in which we have, at 
present, a right to speak of solids being equal, namely in the sense that they 
can be applied, the one to the other. They cannot be so applied because the 
faces, though equal respectively, are not similarly arranged ; consequently the 
prisms are symmetrical, and it ought to be proved that they are, though not 
equal and similar, equal in content, or equivalent, as Legendre has it. 

Legendre addressed himself to proving that the two prisms are equivalent, 
and his method has been adopted, though his 
name is not mentioned, by Schultze and Seven- 
oak and by Holgate. Certain preliminary pro- 
positions are necessary. 

1. The sections of a prism made by parallel 
planes cutting all the lateral edges art equal 
polygons. 

Suppose a prism MNcut by parallel planes 
which make sections ABCDE, A'BCDE. 

1iowAB,BC, CD, . . . are respectively parallel 
to A'B 1 , BC, CD',.... [xt. 16] 

Therefore the angles ABC, BCD, ... are 
equal to the angles A'BC, BCD, ... respec- 
tively, [xi. 10] 

Also AB, BC, CD, ... are respectively equal 
to A'B, BC, CD 1 ,.... [1.34] 

Thus the polygons ABCDE, A'BCDE' are equilateral and equiangular 
to one another. 

1. Two prisms are equal when they have a solid angle in each contained by 
three faces equal each to each and similarly arranged. 

Let the faces ABCDE, AG, AL be equal and similarly placed to the 
faces A'BCDE, A'G", A'L'. 

Since the three plane angles at A, A' are equal respectively and are 
similarly placed, the trihedral angle at A is equal to the trihedral angle at A'. 

[(3) in note to xi. 21] 




33* 



BOOK XI 



Place the trihedral angle at A on that at A'. 
Then the face ABCDE coincides with the face A'ffC'DE, the 
with the face A'G\ and the face AL with the face A'L'. 
The point C falls on C" and D on D. 



[xi. 28 
lace v4£7 







Since the lateral edges of a prism are parallel, CH will fall an CJT, and 
DKonDK'. 

And the points -^ G, L coincide respectively with F, G', L', so that 
the planes GfC, G'K' coincide. 

Hence H, K coincide with H', K' respectively. 

Thus the prisms coincide throughout and are equal. 

In the same way we can prove that two truncated prisms with three faces 
forming a solid angle related to one another as in the above proposition are 
identically equal. 

In particular, 

Cor. Two right prisms having equal bases and equal heights are equal. 

3. An oblique prism is equivalent to a right prism whose base is a right 
section of the oblique prism and whose 
height is equal to a lateral edge of the 
oblique prism. 

Suppose GL to be a right section of 
the oblique prism AD, and let GL be 
a right prism on GL as base and with 
height equal to a lateral edge of AD. 

Now the lateral edges of GL are 
equal to the lateral edges of AD', 

Therefore AG=A'G\ BH=BH', 
CK= C'/C, etc. 

Thus the faces AH, BK, CL are 
equal respectively to the faces A' IT, 
BK 1 , CL'. 

Therefore [by the proposition 
above] 

(truncated prism AL) = (truncated 

prism A'L'). 
Subtracting each from the whole solid AL', we see that 
the prisms AD, GL' are equivalent. 




xi. i8, 29] 



PROPOSITIONS 28, 29 



333 



Now suppose the parallelepiped of Euclid's proposition to be cut by the 
plane through AG, DF 

Let KLMNhe. a right section of the parallelepiped 
cutting the edges AD, BC, GF, HE. 

Then KLMN is a parallelogram; and, if the 
diagonal KM be drawn, 

C±KLM=&MNK. 

Now the prism of which the As ABG, DCF are 
the bases is equal to the right prism on A KLM as 
base and of height AD. 

Similarly the prism of which the As AGH, DFE 
are the bases is equal to the right prism on &MNK 
as base and with height AD. [(3) above] 

And the right prisms on A s KLM, MNK as bases and of equal height 
AD are equal. [(*), Cor. above] 

Consequently the two prisms into which the parallelepiped is divided are 
equivalent. 




Proposition 29, 

Parallelepipedal solids wkich are on the same base and of 
the same height, and in wkich the extremities of the sides which 
stand up are on the same straight lines, are equal to one 
another. 

Let CM, CN be parallelepipedal solids on the same base 
AB and of the same height, 
and let the extremities of their 
sides which stand up, namely 
AG, AF, LM, LN, CD, CE, 
BH,BJC,be on the same straight 
lines FN,DK; 

I say that the solid CM is equal 
to the solid CN. 

For, since each of the figures 
CH, CK is a parallelogram, CB 
is equal to each of the straight lines DH, EK , 

hence DH is also equal to EK. 

Let EH be subtracted from each ; 
therefore the remainder DE is equal to the remainder HK. 

Hence the triangle DCE is also equal to the triangle 
HBK, [i 8, 4 ] 

and the parallelogram DG to the parallelogram HN. [1. 3°] 



. 1 


t 


H 


K 


rV 


d 


«y\ 




N 


Nl 


\ 


1 



[I- 34] 



334 BOOK XI [xl %% 30 

For the same reason 
the triangle AFG is also equal to the triangle MLN. 

But the parallelogram CF is equal to the parallelogram BM, 
and CG to BN, for they are opposite ; 

therefore the prism contained by the two triangles AFG, DCE 
and the three parallelograms AD, DG, CG is equal to the 
prism contained by the two triangles MLN, HBK and the 
three parallelograms BM, HN, BN, 

Let there be added to each the solid of which the 
parallelogram AB is the base and GEHM its opposite ; 
therefore the whole parallelepipedal solid CM is equal to the 
whole parallelepipedal solid CN. 



Therefore etc. 



Q. E. D. 



As usual, Euclid takes one case only and leaves the reader to prove for 
himself the two other possible cases shown in the subjoined figures. Euclid's 
proof holds with a very slight change in each case. With the first figure, the 




only difference is that the prism of which the As GAL, ECB are the bases 
takes the place of " the solid of which the parallelogram AB is the base and 
GEHM its opposite"; while with the second figure we have to subtract the 
prisms which are proved equal successively from the solid of which the 
parallelogram A 3 is the base and FDKN its opposite. 

Simson, as usual, suspects mutilation by "some unskilful editor," but gives 
a curious reason why the case in which the two parallelograms opposite to 
AB have a side common ought not to have been omitted, namely that this 
case "is immediately deduced from the preceding 18th Prop which seems for 
this purpose to have been premised to the 19th." But, apart from the fact that 
Euclid's Prop. 28 does not prove the theorem which it enunciates (as we have 
seen), that theorem is not in the least necessary for the proof of this case of 
Prop. 29, as Euclid's proof applies to it perfectly well. 

Proposition 30. 

Parallelepipedal solids which are on the same base and of 
ike same keigkl, and in which ike extremities of the sides wkuk 
stand up are not on the same straight lines, are equal to one 
another. 



XI. 3°] 



PROPOSITIONS 39, 30 



335 




Let CM, CN be parallelepipedal solids on the same base 
AB and of the same height, 
and lettheextremitiesoftheir 
sides which stand up, namely 
AF,AG,LM,LN,CD,CE, 
BH t BK, not be on the same 
straight lines ; 

I say that the solid CM is 
equal to the solid CN, 

For let NK, DH be pro- 
duced and meet one another 
at R, 

and further let FM, GE be 
produced to P, Q ; 
let AO, LP, CQ, BR be joined. 

Then the solid CM, of which the parallelogram ACBL is 
the base, and FDHM its opposite, is equal to the solid CP, 
of which the parallelogram ACBL is the base, and OQRP its 
opposite ; 

for they are on the same base A CBL and of the same height, 
and the extremities of their sides which stand up, namely AF t 
AO, LM, LP, CD, CQ, BH, BR, are on the same straight 
lines FP, DR. [xi. 39] 

But the solid CP, of which the parallelogram ACBL is 
the base, and OQRP its opposite, is equal to the solid CN, 
of which the parallelogram ACBL is the base and GEKN its 
opposite ; 

for they are again on the same base A CBL and of the same 
height, and the extremities of their sides which stand up, 
namely AG, AO, CE, CQ, LN, LP, BK. BR, are on the 
same straight lines GQ, NR. 

Hence the solid CM is also equal to the solid CN. 
Therefore etc. 

Q. E. D. 
This proposition completes the proof of the theorem that 
Tkoo parallelepipeds on the same base and of the same height are equivalent. 
Legendte deduced the useful theorem that 

Every parallelepiped can be changed into an equivalent rectangular parallele- 
piped having the same iteight and an equivalent base. 

For suppose we have a parallelepiped on the base ABCD with EFGH for 
the opposite face. 



35* 



BOOK XI 



[xh 30 



Draw A I, BK, CL, DM perpendicular to the plane through EFGH and 
all equal to the height of the parallelepiped AG. Then, on joining IK, KL, 
LM, MI, we have a parallelepiped equivalent to the original one and having 
its lateral faces AK, BL, CM, Dl rectangles. 



-^ 




If ABCD is not a rectangle, draw AO, DN in the plane AC perpendicu- 
lar to BC, and IP, MQ in the plane IL perpendicular to KL. 

Joining OP, NQ, we have a rtdangular parallelepiped on AOND as base 
which is equivalent to the parallelepiped with ABCD as ba 1 ^ and IKLM as 
opposite face, since we may regard these parallelepipeds as being on the same 
base ADM J 'and of the same height (AO). 

That is, a rectangular parallelepiped has been constructed which is 
equivalent to the given parallelepiped and has (1) the same height, (2) an 
equivalent base. 

The American text-books which I have quoted adopt a somewhat different 
construction shown in the subjoined figure. 




The edges AB, DC, EF, HG of the original parallelepiped are produced 
and cut at right angles by two parallel planes at a distance apart A B equal 
to AB. 

Thus a parallelepiped is formed in which all the faces are rectangles except 
A' IT, BG\ 



xi. 3°. 3i] PROPOSITIONS 30, 31 337 

Next produce I/A', CB, G'F, H'E 'and cut them perpendicularly by two 
parallel planes at a distance apart B'C equai to BC. 

The points of section determine a rectangular parallelepiped. 

The equivalence of the three parallelepipeds is proved, not by EucL xi. 
*9> 30, but by the proposition about a right section of a prism given above in 
the note to xi. 18 (3 in that note). 



Proposition 31. 

Parallelepipedal solids which are on equal bases and of the 
same height are equal to one another. 

Let the parallelepipedal solids AE, CF, of the same height, 
be on equal bases AB, CD. 

I say that the solid AE is equal to the solid CF. 
























First, let the sides which stand up, HK, BE, AG, LJIf, 
PQ, DF, CO, RS, be at right angles to the bases AB, CD ; 
let the straight line RT be produced in a straight line 
with CR\ 

on the straight line RT, and at the point R on it, let the 
angle TR U be constructed equal to the angle ALB, [1. 23] 
let RT be made equal to AL, and RU equal to LB, 
and let the base ^f^and the solid XC/he completed. 

Now, since the two sides TR, RU are equal to the two 
sides AL, LB, 

and they contain equal angles, 

therefore the parallelogram R W is equal and similar to the 

parallelogram HL. 

Since again AL is equal to RT, and LM to RS, 
and they contain right angles, 



33* BOOK XI [xi. 31 

therefore the parallelogram RX is equal and similar to the 
parallelogram AM. 

For the same reason 
LE is also equal and similar to SU \ 

therefore three parallelograms of the solid AE are equal and 
similar to three parallelograms of the solid XU. 

But the former three are equal and similar to the three 
opposite, and the latter three to the three opposite ; [xi. 24] 
therefore the whole parallelepipedal solid AE is equal to the 
whole parallelepipedal solid XU. [xt Def. 10] 

Let DR, WU be drawn through and meet one another 
at Y, 

let a Tb be drawn through T parallel to D Y, 
let PD be produced to a, 
and let the solids YX, RI be completed. 

Then the solid XY, of which the parallelogram RX is the 
base and Yc its opposite, is equal to the solid XU of which 
the parallelogram RX is the base and UV its opposite, 
for they are on the same base RX and of the same height, and 
the extremities of their sides which stand up, namely RY, RU, 
Tb, TW, Se t Sd, Xc, XV, are on the same straight lines 
YW,eV. ' [xi. 29] 

But the solid XU is equal to AE : 
therefore the solid XY is also equal to the solid AE. 

And, since the parallelogram RUWT is equal to the 
parallelogram YT 

for they are on the same base RT and in the same parallels 
RT, YW, [.. 35] 

while R UWT is equal to CD, since it is also equal to AB, 
therefore the parallelogram YT is also equal to CD. 

But DTis another parallelogram ; 
therefore, as the base CD is to D T, so is YT to D T. [v. 7] 

And, since the parallelepipedal solid CI has been cut by 
the plane RF which is parallel to opposite planes, 
as the base CD is to the base DT, so is the solid CF to the 
solid RI. [xi. a S ] 



xi. 3 i] 



PROPOSITION 31 



339 



For the same reason, 

since the parall el epi pedal solid YI has been cut by the plane 
RX which is parallel to opposite planes, 

as the base YT is to the base TD, so is the solid YX to the 
solid RI. [xt 25] 

But, as the base CD is to DT, so is K^to DT; 

therefore also, as the solid CF is to the solid RI, so is the 
solid YX to RI. [v. n] 

Therefore each of the solids CF, YX has to RI the same 
ratio ; 
therefore the solid CF is equal to the solid YX. [v. 9] 

But YX was proved equal to AE ; 
therefore AE is also equal to CF. 

Next, let the sides standing up, AG, HK, BE, LM, CN, 
PQ, DF, RS, not be at right angles to the bases AB, CD ; 

I say again that the solid AE is equal to the solid CF. 

Q F 





For from the points K, E, G, M, Q, F, N, S let KO, ET, 
GU,MV, QW, FX, NY, SI be drawn perpendicular to the 
plane of reference, and let them meet the plane at the points 
0, T, U, V W, X, Y, I, 
and let OT, OU, UV, TV, WX, WY, YI, IX be joined. 

Then the solid KV'vs, equal to the solid QI, 

for they are on the equal bases KM, QS and of the same 
height, and their sides which stand up are at right angles to 
their bases. [First part of this Prop.] 

But the solid KV is equal to the solid AE, 
and QI to CF; 

for they are on the same base and of the same height, while 
the extremities of their sides which stand up are not on the 
same straight lines. [3". 3°] 



34° BOOK XI [xi. 31 

Therefore the solid AE is also equal to the solid CF. 
Therefore etc. 

Q. E. D. 

It is interesting to observe that, in the figure of this proposition, the bases 
are represented as lying " in the plane of the paper," as it were, and the third 
dimension as " standing up " from that plane. The figure is that of the 
manuscript P slightly corrected as regards the solid AE. 

Nothing could well be more ingenious than the proof of this proposition, 
which recalls the brilliant proposition t. 44 and the proofs of vi. 14 and 33. 

A3 the proof occupies considerable space in the text, it will no doubt be 
well to give a summary. 

I. First, suppose that the edges terminating at the angular points of the 
bases are perpendicular to the bases. 

AB, CD being the bases, Euclid constructs a solid identically equal to 
AE (he might simply have moved AE itself), placing it so that RS is the edge 
corresponding to HK {RS- HK because the heights are equal), and the face 
RX corresponding to HE is in the plane of CS. 

The faces CD t RW are in one plane because both are perpendicular to 
RS. Thus DR, WV meet, if produced, in Y say. 

Complete the parallelograms YT, £>Tand the solids YX, FT. 

Then (solid YX) = (solid UX), 

because they are on the same base ST and of the same height. [xi. 29] 

Also, CI, YI being parallelepipeds cut by planes RF, RX parallel to pairs 
of opposite faces respectively, 

(solid CF) \ (solid RI)=UCD:U DT, [xi. as] 

and (solid YX) : (solid RI)=C3 YT-.CJDT. 

But [1.35] UYT=aUT 

=UAB 

=CJC£>, by hypothesis. 
Therefore (solid CF) = (solid YX) 

m (solid UX) 
= (solid AE). 

II, If the edges terminating at the base are net perpendicular to it, turn 
each solid into an equivalent one on the same base with edges perpendicular 
to it (by drawing four perpendiculars from the angular points of the base to 
the plane of the opposite face), (xi. 29, 30 prove the equivalence.) 

Then the equivalent solids are equal, by Part 1. ; so that the original solids 
are also equal. 

Simson observes that Euclid has made no mention of the case in which 
the bases of the two solids are equiangular, and he prefixes this case to Part L 
in the text. This is surety unnecessary, as Part 1. covers it well enough : the 
only difference in the figure is that UW would coincide with Yb and dV 
with et. 

Simson further remarks that in the demonstration of Part 11. it is not 
proved that the new solids constructed in the manner described art parallele- 
pipeds. The proof is, however, so simple that it scarcely needed insertion 



XL 31, 3*] PROPOSITIONS 31, 3 a 341 

into the text He is correct in his remark that the words "while the 
extremities of their sides which stand up are not on the same straight lines " 
just before the end of the proposition would be better absent, since they may 
be " on the same straight lines." 

Proposition 32. 

Parallelepipedal solids which are of the same height are to 
one another as their hoses. 

Let AB, CD be parallelepipedal solids of the same height ; 
I say that the parallelepipedal solids AB, CD are to one 
another as their bases, that is, that, as the base AE is to the 
base CF, so is the solid AB to the solid CD, 




For let FH equal to AE be applied to FG, [1. 45] 

and on FH as base, and with the same height as that of CD, 
let the parallelepipedal solid GK be completed. 

Then the solid AB is equal to the solid GK ; 
for they are on equal bases AE, FH and of the same height. 

[xi. 31] 

And, since the parallelepipedal solid CK is cut by the plane 
DG which is parallel to opposite planes, 
therefore, as the base CF is to the base FH, so is the solid 
CD to the solid DH. [xi. 25] 

But the base FH is equal to the base AE, 
and the solid GK to the solid AB ; 

therefore also, as the base AE is to the base CF, so is the 
solid AB to the solid CD. 

Therefore etc. 

Q. E. D. 

As Clavius observed, Euclid should have said, in applying the parallelo- 
gram FH to FG, that it should be applied " in the angle FGH equal to the 
angle LCG." Simson is however, I think, hypercritical when he states as 
regards the completion of the solid GK that it ought to be said, " complete 



34* 



BOOK XI 



[xi- 3*. 33 



the solid of which the base is FH, and one of its insisting straight lines is FD." 
Surely, when we have two faces DG, FH meeting in an edge, to say "complete 
the solid " is quite sufficient, though the words " on FH as base " might 
perhaps as well be left out. The same " completion " of a paralletepipedal 
solid occurs in xi. 31 and 33. 



Proposition 33. 

Similar parallekpipedal solids are to one another in the 
triplicate ratio of their corresponding sides. 

Let AB, CD be similar parallelepipedal solids, 
and let AE be the side corresponding to CF; 
I say that the solid AB has to the solid CD the ratio triplicate 
of that which AE has to CF. 











For let EK, EL, EM be produced in a straight line with 
AE, GE, HE, 

let EK be made equal to CF, EL equal to FN, and further 

EM equal to FR, 

and let the parallelogram KL and the solid KP be completed. 

Now, since the two sides KE, EL are equal to the two 
sides CF, FN, 

while the angle KEL is also equal to the angle CFN, 
inasmuch as the angle AEG is also equal to the angle CFN 
because of the similarity of the solids AB, CD, 



XX. 33] PROPOSITIONS 3 », 33 343 

therefore the parallelogram KL is equal < and similar > to the 
parallelogram CN. 

For the same reason 
the parallelogram KM is also equal and similar to CR, 
and further EP to DF; 

therefore three parallelograms of the solid KP are equal and 
similar to three parallelograms of the solid CD. 

But the former three parallelograms are equal and similar 
to their opposites, and the latter three to their opposites ; [xi. 24] 
therefore the whole solid KP is equal and similar to the whole 
solid CD, [xi. Def. ro] 

Let the parallelogram GK be completed, 
and on the parallelograms GK, KL as bases, and with the 
same height as that of AB, let the solids EO, LQ be 
completed. 

Then since; owing to the similarity of the solids AB, CD, 
as AB is to CF, so is EG to FN, and EH to FR, 
while CF is equal to EK, FN to EL, and FR to EM, 
therefore, as AE is to EK, so is GE to EL, and HE to EM. 

But, as AE is to EK, so is A G to the parallelogram GK, 
as GE is to EL, so is GK to KL, 

and, as HE is to EM, so is QE to ATAf ; [?i. 1] 

therefore also, as the parallelogram AG is to GK, so is GK 
to AX, and QE to A'AT. 

But, as A G is to GK, so is the solid AB to the solid Zj 0, 
as GK is to A"Z, so is the solid OE to the solid QL, 
and, as QE is to AT./B/, so is the solid QL to the solid KP ; 

[xi. 31] 
therefore also, as the solid AB is to EO, so is EO to QL, and 
0£ to AV. 

But, if four magnitudes be continuously proportional, the 
first has to the fourth the ratio triplicate of that which it has 
to the second ; [v. Def. 10] 

therefore the solid AB has to KP the ratio triplicate of that 
which AB has to EO. 

But, as AB is to EO, so is the parallelogram AG to GK, 
and the straight line AE to EK [vi. 1] ; 



344 BOOK XT [xi. 33 

hence the solid AB has also to KP the ratio triplicate of that 
which AE has to EK. 

But the solid KP is equal to the solid CD, 
and the straight line EK to CF\ 

therefore the solid AB has also to the solid CD the ratio 
triplicate of that which the corresponding side of it, AE, has 
to the corresponding side CF. 

Therefore etc. 

Q. E. D. 

Porism. From this it is manifest that, if four straight 
lines be < continuously > proportional, as the first is to the 
fourth, so will a parallelepipedal solid on the first be to the 
similar and similarly described parallelepipedal solid on the 
second, inasmuch as the first has to the fourth the ratio 
triplicate of that which it has to the second. 

The proof may be summarised as follows. 

The three edges AE, GE, HE of the parallelepiped AB which meet at 
E, the vertex corresponding to R in the other parallelepiped, are produced, 
and lengths EK, EL, EM are marked off equal respectively to the edges CF, 
FN, ER of CD. 

The parallelograms and solids are then completed as shown in the figure. 
Euclid first shows that the solid CD and the new solid PK are equal and 
similar according to the criterion in xi. Def. 10, viz. that they are contained 
by the same number of equal and similar planes. (They are arranged in the 
same order, and it would be easy to prove equality by proving the equality of 
a pair of solid angles and then applying one solid to the other.) 
We have now, by hypothesisj 

AE: CE= EG : EN '= EH ': ER ; 
that is, AE : EK= EG: EL = EH : EM. 

But AE;EK = E3AG:CDGK, [vi. 1] 

EG: EL=E3GK:EJKL, 
EH : EM=n HK : U KM. 
Again, by xi. 25 or 32, 

CJAG-.CJGK^ (solid AB) : (solid EO), 
CJGK-.U KL = (solid EO) : (solid QL), 
CJHK-.CD KM= (solid QL) : (solid KP). 
Therefore 
(solid AB) : (solid EO) = (solid EO) : (solid QL) = (solid QL) : (solid KP), 
or the solid AB is to the solid KP (that is, CD) in the ratio triplicate of that 
which the solid AB has to the solid EO, i.e. the ratio triplicate of that which 
AE has to EK (or CF). 

Heiberg doubts whether the Porism appended to this proposition is 
genuine. 



XI. 3$, 34] PROPOSITIONS 33, 34 345 

Simson adds, as Prop. D, a useful theorem which we should have expected 
to find here, on the analogy of vi. 23 following vi. 19, 20, viz. that Solid 
parallelepipeds contained by parallelograms equiangular to one another, each to 
each, that is, of which the solid angles are equal, each to each, have to one another 
the ratio compounded of the ratios of their sides. 

The proof follows the method of the proposition xi. 33, and we can use 
the same figure. In order to obtain one ratio between lines to represent the 
ratio compounded of the ratios of the sides, after the manner of vi. 13, we 
take any straight lint a, and then determine three other straight lines b, c, d, 
such that 

AE: CF=a:b, 

EG:FN=b;c, 

EH:FR^c:d, 

whence a : d represents the ratio compounded of the ratios of the sides. 

We obtain, in the same manner as above, 

(solid AB) : (solid EO) =C3 AG :U GK= AE : EK= AE : CF 

= a:b, 
(solid EO) : (solid QL)=a GK-.CJ KL = GE : EL = GE : FN 

= b:c, 
(solid QL): (solid KF)=CJ HK:CJKM=EH;EM=EH:FR 

= cid, 
whence, by composition [v. 2a], 

(solid AB) : (solid KP) = a:d, 
or {solid AB) : (solid CD) = a:d. 

Proposition 34. 

In equal parallelepipedal solids the bases are reciprocally 
proportional to the heights; and those parallelepipedal solids in 
which the bases are reciprocally proportional to the heights are 
equal. 

Let AB, CD be equal parallelepipedal solids ; 
I say that in the paralieiepipedal solids AB, CD the bases are 
reciprocally proportional to the heights, 

that is, as the base EH is to the base NQ, so is the height 
of the solid CD to the height of the solid AB. 

First, let the sides which stand up, namely AG, EF, LB, 
HK, CM, NO, PD, QR, be at right angles to their bases • 
I say that, as the base EH is to the base NQ, so is CM 
to AG. 

If now the base EH is equal to the base NQ, 
while the solid AB is also equal to the solid CD, 
CM will also be equal to AG. 



346 



BOOK XI 



[*'■ 34 



For parallelepipedal solids of the same height are to 
one another as the bases ; [xi. 3*] 

and, as the base EH is to NQ, so will CM be to AG, 
and it is manifest that in the parallelepipedal solids AB, CD 
the bases are reciprocally proportional to the heights. 

Next, let the base EH not be equal to the base NQ, 
but let EH be greater. 



! 


< 




1 


,/ 


/ 






H 




L 


/ 




/ 


A 


t 





R 

\ A 

T 


C N 



Now the solid AB is equal to the solid CD ; 
therefore CM is also greater than AG. 

Let then CT be made equal to A G, 
and let the parallelepipedal solid VC be completed on NQ as 
base and with CT as height 

Now, since the solid AB is equal to the solid CD, 
and C V is outside them, 

while equals have to the same the same ratio, [v. 7] 

therefore, as the solid AB is to the solid CV, so is the solid 
CD to the solid CV. 

But, as the solid AB is to the solid CV, so is the base 
EH to the base NQ, 

for the solids AB, CV are of equal height ; [xi. 32] 

and, as the solid CD is to the solid CV,$o is the base MQ to 
the base TQ [xi. 25] and CM to CT [vi. 1] ; 
therefore also, as the base EH is to the base NQ, so is MC 
to CT 

But CT is equal to A G ; 
therefore also, as the base EH is to the base NQ, so is MC 
to AG, 



XX. 34] PROPOSITION 34 347 

Therefore in the para) 1 el epi pedal solids AB, CD the bases 

are reciprocally proportional to the heights. 

Again, in the parallelepipedal solids AB CD let the bases 
be reciprocally proportional to the heights, chat is, as the base 
EH is to the base NQ, so let the height of the solid CD be 
to the height of the solid AB \ 
I say that the solid AB is equal to the solid CD. 

Let the sides which stand up be again at riglit angles to 
the bases. 

Now, if the base EH is equal to the base NQ, 
and, as the base EH is to the base NQ, so is the height of 
the solid CD to the height of the solid AB, 
therefore the height of the solid CD is also equal to the 
height of the solid AB, 

But parallelepipedal solids on equal bases and of the same 
height are equal to one another ; [xi. 31] 

therefore the solid AB is equal to the solid CD. 

Next, let the base EH not be equal to the base NQ, 
but let EH be greater ; 

therefore the height of the solid CD is also greater than the 

height of the solid AB, 

that is, CM is greater than AG. 

Let CT'be again made equal to AG, 
and let the solid CVbe similarly completed. 

Since, as the base EH is to the base NQ, so is MC 
to AG, 

while AG is equal to CT, 

therefore, as the base EH is to the base NQ, so is CM 
to CT. 

But, as the base EH is to the base NQ, so is the solid 
AB to the solid CV, 

for the solids AB, C V are of equal height ; [x jz] 

and, as CM is to CT, so is the base MQ to the base Q T [n. 1] 
and the solid CD to the solid CV. [ax as] 

Therefore also, as the solid AB is to the solid CV, so is 
the solid CD to the solid CV; 

therefore each of the solids AB, CD has to CJ' the same 
ratio. 



348 BOOK XI [xi. 34 

Therefore the solid AB is equal to the solid CD. [v. 9] 

Now let the sides which stand up, FE, BL, GA, HK, 
ON, DP, MC, RQ, not be at right angles to their bases ; 
let perpendiculars be drawn from the points F, G, B, K, 0, 
M, D, R to the planes through EH, NQ, and let them meet 
the planes at S, T, U, V, W, X, Y, a, 
and let the solids FV, Oa be completed ; 
I say that, in this case too, if the solids AB, CD are equal, 
the bases are reciprocally proportional to the heights, that is, 
as the base EH is to the base NQ, so is the height of the 
solid CD to the height of the solid AB. 











Since the solid AB is equal to the solid CD, 
while AB is equal to BT, 

for they ate on the same base FK and of the same height; 

[xi. 29, 30] 
and the solid CD is equal to DX, 

for they are again on the same base RO and of the same 
height ; [#•] 

therefore the solid BT is also equal to the solid DX. 

Therefore, as the base FK is to the base OR, so is the 
height of the solid DX to the height of the solid BT. 

[Part 1.] 

But the base FK is equal to the base EH, 
and the base OR to the base NQ ; 

therefore, as the base EH is to the base NQ, so is the height 
of the solid DX to the height of the solid BT. 



xi. 34] PROPOSITION 34 349 

But the solids DX, BT and the solids DC, BA have the 
same heights respectively ; 

therefore, as the base EH is to the base NQ, so is the height 
of the solid DC to the height of the solid AB, 

Therefore in the parallelepipedal solids AB, CD the bases 
are reciprocally proportional to the heights. 

Again, in the parallelepipedal solids AB, CD let the bases 
be reciprocally proportional to the heights, 
that is, as the base EH is to the base NQ, so let the height 
of the solid CD be to the height of the solid AB ; 
I say that the solid AB is equal to the solid CD. 

For, with the same construction, 
since, as the base EH is to the base NQ, so is the height of 
the solid CD to the height of the solid AB, 
while the base EH is equal to the base FK, 
and NQ to OR, 

therefore, as the base FK is to the base OR, so is the height 
of the solid CD to the height of the solid AB. 

But the solids AB, CD and BT, DX have the same 
heights respectively ; 

therefore, as the base FK is to the base OR, so is the height 
of the solid DX to the height of the solid BT. 

Therefore in the parallelepipedal solids BT, DX the bases 
are reciprocally proportional to the heights ; 
therefore the solid BT is equal to the solid DX. [Part 1.] 

But BTis equal to BA, 

for they are on the same base FK and of the same height ; 

[xi. 29, 30] 
and the solid DX is equal to the solid DC. [id.] 

Therefore the solid AB is also equal to the solid CD. 

Q. E. D. 

In this proposition Euclid makes two assumptions which require notice, 
(j) that, if two parallelepipeds are equal, and have equal bases, their heights 
are equal, and (2) that, if the bases of two equal parallelepipeds are unequal, 
that which has the lesser base has the greater height In justification of the 
former statement Euclid says, according to what Heiberg holds to be the 
genuine reading, " for parallelepipedal solids of the same height are to one 
another as their bases" [xi. 3*]. This apparently struck some very early 
editor as not being sufficient, and he added the explanation appearing in 
Sim son's text, " For if, the bases EH, NQ being equal, the heights AG, CM 



3 So BOOK XI [xi. 34 

were not equal, neither would the solid AB be equal to CD. But it is by 
hypothesis equal. Therefore the height CM is not unequal to the height AG; 
therefore it is equal." Then, it being perceived that there ought not to be two 
explanations, the genuine one was erased from the inferior mss. While the 
interpolated explanation does not take us very far, the truth of the statement 
may be deduced with perhaps greater case from xi. 31 than from xi. 32 
quoted by Euclid. For, assuming one height greater than the other, while the 
bases are equal, we have only to cut from the higher solid so much as will 
make its height equal to that of the other. Then this part of the higher solid 
is equal to the whole of the other solid which is by hypothesis equal to the 
higher solid itself. That is, the whole is equal to its part ; which is impossible. 

The genuine text contains no explanation of the second assumption that, 
if the base EH be greater than the base NQ, while the solids are equal, the 
height CM is greater than the height AG; for the added words "for, if not, 
neither again will the solids AB, CD be equal ; but they are equal by 
hypothesis " are no doubt interpolated. In this case the truth of the assump- 
tion is easily deduced from xi. 32 by redudie ad absurdum. If the height CM 
were equal to the height AG, the solid AB would be to the solid CD as the 
base EH is to the base NQ, i.e. as a greater to a less, so that the solids would 
not be equal, as they are by hypothesis. Again, if the height CM were less 
than the height AG, we could increase the height of CD till it was equal to 
that of AB, and it would then appear that AB is greater than the heightened 
solid and a fortiori greater than CD : which contradicts the hypothesis. 

Clavius rather ingeniously puts the first assumption the other way, saying 
that, if the heights are equal in the equal parallelepipeds, the bases must be 
equal This follows directly from xi. 32, which proves that the parallelepipeds 
are to one another as their bases ; though Clavius deduces it indirectly from 
xi. 31. The advantage of Clavius' alternative is that it makes the second 
assumption unnecessary. He merely says, if the heights be not equal, let CM 
be the greater, and then proceeds with Euclid's construction. 

It is also to be observed that, when Euclid comes to the corresponding 
proposition for cones and cylinders [xn. 15 J he begins by supposing the 
heights equal, inferring by HI. 11 (corresponding to xi. 31) that, the solids 
being equal, the bases are also equal, and then proceeds to the case where the 
heights are unequal without making any preliminary inference about the 
bases. The analogy then of xn, 1 5, and the fact that he quotes xi. 3* here 
(which directly proves that, if the solids are equal, and also their heights, their 
bases are also equal), make Clavius' form the more convenient to adopt. 

The two assumptions being proved as above, the proposition can be put 
shortly as follows. 

I. Suppose the edges terminating at the comers of the base to be per- 
pendicular to it. 

Then (a), if the base EH be equal to the base NQ, the parallelepipeds 
being also equal, the heights must be equal (converse of xi. 31), so that the 
bases are reciprocally proportional to the heights, the ratio of the bases and 
the ratio of the heights being both ratios of equality. 

(£) If the base EH be greater than the base NQ, and consequently (by 
deduction from xi. 32) the height CM greater than the height AG, cut off 
CT from CM equal to AG, and draw the plane TV through T parallel to the 
base NQ, making the parallelepiped CV, with CT{= AG) for its height. 

Then, since the solids AB, CD are equal, 

(solid AB) : (solid CV) = (solid CD) : (solid CV). [v. 7] 



xi. 34] PROPOSITION 34 351 

Bui (solid AB) : (solid CV)=C3 HE -.a NQ, [xi. 32] 

and (solid CD) : (solid C V) = O MQ : O TQ [xi. 35] 

= CM:CT. [vi. 1 J 

Therefore E3 HE-.U NQ=CM;CT 

= CM:AG. 
Conversely («), if the bases EH, NQ be equal and reciprocally proportional 
to the heights, the heights must be equal. 

Consequently (solid AB) = (solid CD), [xi. 31] 

{») If the bases EH, NQ be unequal, if, e.g. O EH>£JNQ, 
then, since O -ff-ff : CD NQ = CjV : A G, 

CM>AG. 
Make the same construction as before. 

Then CJEH:ONQ = (solid AB) : (solid CV\ [xi. 32] 

and CM:AG=CM: CT 

= nDMQ;C]TQ [vi. 1] 

= (solid CD) ■ (solid CF). [xi. 25] 

Therefore 

(solid AB) : (solid CV) = (solid CZ>) : (solid CF), 
whence (solid -4 J) = solid CD. [v. 9] 

II. Suppose that the edges terminating at the corners of the bases are not 
perpendicular to it 

Drop perpendiculars on the bases from the corners of the faces opposite 
to the bases. 

We thus have two parallelepipeds equal to AB, CD respectively, since 
they are on the same bases FK, RO and of the same height respectively. 

[xi. so, 30] 
If then (1) the solid AB is equal to the solid CD, 
(solid BT) = (solid i?*), 
and, by the first part of this proposition, 

O KF : a OB = MX : GT, 

or £7 HE :0 NQ = MX : GT. 

(a) If SDHE-.aNQ^MX-.GT, 

then o KF: CJOR = MX : GT, 

so that, by the first half of the proposition, the solids BT, DX are equal, and 
consequently 

(solid AB) m (solid CD). 



The text of the second part of the proposition four times contains, after 
the words " of the same height," the words " in which the sides which stand 
up are not on the same straight lines." As Sim son observed, they are inept, 
as the extremities of the edges may or may not be " on the same straight 
lines"; cf. the similar words incorrectly inserted at the end of XI. 31. 

Words purporting to quote the result of the first part of the proposition 
are also twice inserted; but they are rejected as unnecessary and as containing 
an absurd expression — "(solids) in which the heights are at right angles to their 
bases," as if the heights could be otherwise than perpendicular to the bases. 



35* BOOK XI [xi. 35 



Proposition 35. 

If there be two equal plane ^angles, and on their vertices 
there be set up elevated straight lines containing equal angles 
with the original straight lines respectively, if on the elevated 
straight lines points be taken at random and perpendiculars be 
drawn from them to the planes in which the original angles 
are, and if from the points so arising in the planes straight 
lines be joined to the vertices of the original angles, they will 
contain, with the elevated straight lines, equal angles. 

Let the angles BA C, EDF be two equal rectilineal angles, 
and from the points A, D let the elevated straight lines AG, 
DM be set up containing, with the original straight lines, 
equal angles respectively, namely, the angle MDE to the 
angle GAB and the angle MDF to the angle GAC, 
let points G, M be taken at random on AG, DM, 
let GL, MN be drawn from the points G, M perpendicular to 
the planes through BA, AC and ED, DF, and let them meet 
the planes at L, N, 
and let LA, ND be joined ; 
I say that the angle GAL is equal to the angle MDN. 





Let AH be made equal to DM, 
and let HK\x, drawn through the point H parallel to GL. 

But GL is perpendicular to the plane through BA, AC ; 
therefore HK is also perpendicular to the plane through, 
BA, AC. [x., 8] 

From the points K, N let KC, NF, KB, NE be drawn 
perpendicular to the straight lines AC, DF, A3, DE, 
and let HC, CB, MF, FE be joined. 



xi. 35] PROPOSITION 35 353 

Since the square on HA is equal to the squares on HK, 
KA, 

and the squares on KC, CA are equal to the square on KA, 

['■ 47] 
therefore the square on HA is also equal to the squares on 
HK, KC, CA. 

But the square on HC is equal to the squares on 
HK, KC; [1.47] 

therefore the square on HA is equal to the squares on 
HC, CA. 

Therefore the angle HCA is right. [1. 48 J 

For the same reason 
the angle DFM is also right. 

Therefore the angle A CH is equal to the angle DFM. 

But the angle HAC is also equal to the angle MDF. 

Therefore MDF t HAC are two triangles which have two 
angles equal to two angles respectively, and one side equal to 
one side, namely, that subtending one of the equal angles, 
that is, HA equal to MD \ 

therefore they will also have the remaining sides equal to the 
remaining sides respectively. [1. 26] 

Therefore AC is equal to DF. 

Similarly we can prove that AB is also equal to DE, 

Since then AC is equal to DF, and AB to DE, 
the two sides CA, AB are equal to the two sides FD, DE. 

But the angle CAB is also equal to the angle FDE ; 
therefore the base BC is equal to the base EF t the triangle to 
the triangle, and the remaining angles to the remaining 
angles ; [1. 4] 

therefore the angle ACB is equal to the angle DFE. 

But the right angle ACK is also equal to the right angle 
DFN; 

therefore the remaining angle BCK is also equal to the 
remaining angle EFN. 

For the same reason 
the angle CBK is also equal to the angle FEN. 



354 HOOK XI [xi. 35 

Therefore BCK, EFN are two triangles which have two 
angles equal to two angles respectively, and one side equal to 
one side, namely, that adjacent to the equal angles, that is, 
BC equal to EF; 

therefore they will also have the remaining sides equal to the 
remaining sides. [i. *6] 

Therefore CK is equal to FN. 

But AC is also equal to DF ; 
therefore the two sides AC, CK are equal to the two sides 
DF, FN; 
and they contain right angles. 

Therefore the base AK\s equal to the base DN. [i. 4] 

And. since AH is equal to DM, 
the square on AH is also equal to the square on DM, 

But the squares on AK, KH are equal to the square 
on AH, 

for the angle AKH is right ; [1. 47] 

and the squares on DN, NM are equal to the square 
on DM, 

for the angle DNM is right ; [1. 47] 

therefore the squares on AK, KH are equal to the squares 
on DN, NM ; 

and of these the square on A K is equal to the square on DN; 
therefore the remaining square on KH is equal to the square 
on NM; 
therefore HK is equal to MN. 

And, since the two sides HA, AK are equal to the two 
sides MD, DN respectively, 

and the base HK was proved equal to the base MN, 
therefore the angle HAK is equal to the angle MDN. [1. 8] 

Therefore etc. 

Porism. From this it is manifest that, if there be two 
equal plane angles, and if there be set up on them elevated 
straight lines which are equal and contain equal angles with 
the original straight lines respectively, the perpendiculars 
drawn from their extremities to the planes in which are 
the original angles are equal to one another. 

Q. E. D. 



a 35] PROPOSITION 35 355 

This proposition is required for the next, where it is necessary to know 
that, if in two equiangular parallelepipeds equal angles, one in each, be 
contained by three plane angles respectively, one of which is an angle of the 
parallelogram forming the base in one parallelepiped, while its equal is likewise 
in the base of the other, and the edges in which the two remaining angles 
forming the solid angles meet are equal, the parallelepipeds are of the same 
height. 

Bearing in mind the definition of the inclination of a straight line to a 
plane, we might enunciate the proposition more shortly thus. 

If there be two trihedral angles identically equal to one another, corresponding 
edges in each are equally inclined to the planes through the other two edges 
respectively. 

The proof, which is necessarily somewhat long, may be summarised thus. 

It is required to prove that the angles GAL, MDN in the figure are equal, 
G, M being any points on AG, DM, and GL, MN perpendicular to the 
planes BAC, EDF respectively. 

If AH \s made equal to DM, and HK is drawn in the plane GAL parallel 
toCZ, 

HK is also perpendicular to the plane BAC. [xi. 8] 

Draw KB, KC perpendicular to AB, AC respectively and NE, NF 
perpendicular to DE, irrespectively, and complete the figures, 

Now(i) HA* = HFC + KA* 

= HK* + KC* + CA> ■ [i. 47] 

= HC + CA* ) 

Therefore l HCA = a right angle. 

Similarly l. MFD m a right angle. 

(a) As If AC, MDFha.ve therefore two angles equal and one side. 

Therefore AHACsAMDF, and AC=DF. [i. *6] 

(3) Similarly AHAB s AMDE, and AB - DE. 

{4) Hence as ABC, DEF&re equal in all respects, so that BC = EF, 
and lABC=lDEF, 

uACB^^DFE. 

(5) Therefore the complements of these angles are equal, 
i.e. lKBC=lNEF, 

and l.KCB = l.NFE. 

(6) The A s KBC, NEF have two angles equal and one side, and are 
therefore equal in all respects, so that 

KB = NE, 

KC=NF. 

(7) The right-angled triangles KAC, NDFiat equal in all respects, since 
A C= DF[{2) above], KC=NF. 

Consequently AK=DN. 

(8) In As HAK, MDN, 

HK* + KA* = HA* 

= ML?, by hypothesis, 
=-MN* + ND'. 



35° BOOK XI [xl 35, 36 

Subtracting the equals KA\ NIP, 
we have HK* = MN\ 

or HK= MN. 

(9) As HAK, MDNzxt, now equal in all respects, by 1. 8 and 1. 4, and 
therefore 

lHAK=lMDN. 

The Porism is merely a statement of the result arrived at in (8). 

Lege rid re uses, practically, the construction and argument of this propo- 
sition to prove the theorem given under (3) of the note on xi. 21 above that 
In two equal trihedral angles, corresponding pairs of fate angles include equal 
dihedral angles. This fact is readily deduced from the above proposition. 

Since [{1)] HC, KC are both perpendicular to AC, and MF, NF both 
perpendicular to DF, the angles HCK, MFN are the measures of the 
dihedral angles between the planes HAC, BAC, and MDF, EDF respec- 
tively, [xi. Def. 6] 

By (6), KC=NF, 

and, by (8), HK= MN, 

while the angles HKC, MNF, both being right, are equal. 

Consequently the A s HCK, MFN&k equal in all respects, [1. 4] 

so that lHCK=lMFN 

Sim son substituted a different proof of (i) in the above summary, as 
follows. 

Since HK is perpendicular to the plane BAC, the plane If BK, passing 
through HK, is also perpendicular to the plane BAC. [xi. 18] 

And AB, being drawn in the plane BAC perpendicular to BK, the 
common section of the planes HBK, BAC, is perpendicular to the plane 
HBK [xi. Def. 4], and is therefore perpendicular to every straight line 
meeting it in that plane [xi. Def. 3]. 

Hence the angle ABH'xs a right angle. 

I think Euclid's proof much preferable to this with its references to 
definitions which are more of the nature of theorems. 

Proposition 36. 

ff three straight lines be proportional, the parallelepipedal 
solid formed out of the three is equal to the parallelepipedal 
solid on the mean which is equilateral, but equiangular with 
the aforesaid solid. 

Let A, B, C be three straight lines in proportion, so that, 
as A is to B, so is B to C\ 

I say that the solid formed out of A, B, C is equal to the 
solid on B which is equilateral, but equiangular with the 
aforesaid solid. 

Let there be set out the solid angle at E contained by the 
angles DEG, GEF, FED, 



xi. 36] PROPOSITIONS 35, 36 3S7 

let each of the straight lines DE, GE, EF be made equal to 

B, and let the parallelepipedal solid EK be completed, 

let LMhtt made equal to A, 

and on the straight line LM, and at the point L on it, let there 

be constructed a solid angle equal to the solid angle at E, 

namely that contained by NLO, OEM, MLN \ 

let L be made equal to B, and LN equal to C. 




8- 

c- 




Now, since, as A is to B, so is B to C, 
while A is equal to LM, B to each of the straight lines LO, 
ED, and C to LN, 

therefore, as LM is to EF, so is DE to LN, 

Thus the sides about the equal angles NLM, DEE are 
reciprocally proportional ; 

therefore the parallelogram MN is equal to the parallelogram 
DF [vi. 14] 

And, since the angles DEE, NLM are two plane recti- 
lineal angles, and on them the elevated straight lines L O, EG 
are set up which are equal to one another and contain equal 
angles with the original straight lines respectively, 
therefore the perpendiculars drawn from the points G, O to 
the planes through NL, LM and DE, EF are equal to one 
another ; [xi, 35. Por.] 

hence the solids LH, EK are of the same height. 

But parallelepipedal solids on equal bases and of the same 
height are equal to one another ; [xi. 31] 

therefore the solid HL is equal to the solid EK. 

And LH is the solid formed out of A, B, C, and EK the 
solid on B ; 



35* 



BOOK XI 



[» $6, 37 



therefore the parallelepipedal solid formed out of A, B, -C is 
equal to the solid on B which is equilateral, but equiangular 
with the aforesaid solid, 

Q. E. D. 

The edges of the parallelepiped HL being respectively equal to A, B, C, 
and those of the equiangular parallelepiped KE being ail equal to B, we 
regard MN (net containing the edge OL equal to B) as the base of the first 
parallelepiped, and consequently FD, equiangular to MN, as the base of KE. 

Then the solids have the same height [xi. 35, Por.] 

Hence (solid HL) : (solid KE) = O MN : O ED. [xi. 3 2] 

But, since A, B, C are in continued proportion, 
A : B = B : C, 
or LM:EF^DE:LN. 

Thus the sides of the equiangular Di MN, ED are reciprocally pro- 
portional, whence 

UMN=CJFD, [vi. 14] 

and therefore (solid HL) = (solid KE). 



Proposition 37. 

If four straight lines be proportional, the parallelepipedal 
solids on them which are similar and similarly described will 
also be proportional; and, if the parallelepipedal solids on them 
which are similar and similarly described be proportional, the 
straight lines will themselves also be proportional. 

Let AB, CD, EF, GH be four straight lines in proportion, 
so that, as AB is to CD, so is EF to GH ; 
and let there be described on AB, CD, EF, GH the similar 
and similarly situated parallelepipedal solids KA, LC, ME, 
NG; 

I say that, as KA is to LC, so is ME to NG. 



K 



A 




For, since the parallelepipedal solid KA is similar to LC, 
therefore KA has to LC the ratio triplicate of that which AB 
has to CD. [xi. 33] 



x«. 37] 



PROPOSITIONS 36, 37 



For the same reason 

ME also has to NG the ratio triplicate of that which EF has 
to GH, [«] 

And, as AB is to CD, so is EF to GH. 
Therefore also, as AK is to LC, so is ME to NG. 

Next, as the solid AK is to the solid LC, so let the solid 
ME be to the solid NG ; 
I say that, as the straight line AB is to CD, so is EF to <7/£ 

For since, again, KA has to Z.C the ratio triplicate of that 
which AB has to CD, [xi. 33] 

and jfc27T also has to jV(? the ratio triplicate of that which EF 
has to GH, [id.] 

and, as KA is to ZC, so is ME to A^G, 
therefore also, as AB is to CD, so is -ST 7 " to GH. 

Therefore etc. 

Q. E. D. 

In this proposition it is assumed that, if two ratios be equal, the ratio 
triplicate of one is equal to the ratio triplicate of the other and, conversely, 
that, if ratios which are the triplicate of two other ratios are equal, those other 
ratios are themselves equal. 

To avoid the necessity for these assumptions Simson adopts the alternative 
proof found in the MS- which Heiberg calls b, and also adopted by Clavius, 
who, howevet, gives Euclid's proof as well, attributing it to Theon. The 
alternative proof proceeds after the manner of vi. »*, thus. 

Make AB, CD, 0, P continuous proportionals, and also EF, GH, Q, R. 



\ 



i 



D E 

Q 






I. Then, since 

AB:CD = EF: GH, 
we have, tx aequali, 

AB:P=EF:R. [v. «] 

But {solid AK) ■. (solid CL) = AB:P, 

[xi. 33 and For.] ■ 
and (solid EM) : (solid GN) = EF:R. 

Therefore 

(solid AK) \ (solid CL) - (solid EM) : (solid GN) 



360 BOOK XI [xi. 37, 38 

II. If the solids are proportional, take .ST* such that 
AB.CD = EF: ST, 
and on ST describe the parallelepiped SV similar and similarly situated to 
either of the parallelepipeds EM, GN. 

Then, by the first part, 

(solid AK) -. (solid CL) = (solid EM) : (solid SV), 
whence it follows that 

(solid GN) = (solid SV). 

Hut these solids are similar and similarly situated ; 
therefore their faces are similar and equal; [xi. Def. 10] 

therefore the corresponding sides GH, ST are equal. 

[For this inference cf. note on vi. 22. The equality of GH, ST may 
readily be proved by application of the two parallelepipeds to one another, 
since, being similar, they are equiangular.] 

Hence AB : CD = EF. GH. 

The text of the mss. has here a proposition which is as badly placed as it 
is unnecessary. If a plane be at right angles to a plant, and from any one of the 
points in out of the planes a perpendicular be drawn to the other plane, the 
perpendicular so drawn will fall on the. common section of the planes. It is of 
the nature of a lemma to xii. 17, where 
alone the fact is made use of. Heiberg 
observes that it is omitted in b and that the 
copyist of P knew other texts which did not 
contain it, From these facts it is fairly con- 



cluded that the proposition was interpolated. B 

The truth of it is of course immediately 

obvious by rtductio ad absurdum. Let the plane CAD be perpendicular to 
the plane AB, and let a perpendicular be drawn to the latter from any point 
E in the former. 

If it does not fall on AD, the common section, let it meet the plane AB 
inF. 

Draw FG in AB perpendicular to AD, and join EG. 

Then FG is perpendicular to ihe plane CAD[x.\. Def. 4], and therefore 
to GE[x\. Def. 3]. Therefore t, EGF'is right 

Also, since .E^is perpendicular to AB, 
the angle EFG is right 

That is, the triangle EGF has two right angles : 
which is impossible. 

Proposition 38. 

If Ihe sides of the opposite planes of a cube be bisected, and 
planes be carried through the points of section, the common 
section of the planes and the diameter of the cube bisect one 
another. 

For let the sides of the opposite planes CF, AH of the 
cube AF be bisected at the points K, L, -M, N t O, Q, P, R, 



X 



xt. 3 8] 



PROPOSITIONS 37, 38 



361 



and through the points of section let the planes KN t OR be 
carried ; 

let US be the common section of the planes, and DG the 
diameter of the cube AF. 

I say that UT is equal to TS, and DT to TG. 

For let DU, UE, BS, SG be joined. 

Then, since DO is parallel to PE, 

the alternate angles DOU, UPE are equal to one another. 

[1. 29] 
And, since DO is equal to PE, and OU to UP, 
and they contain equal angles, 
therefore the base DU is equal to the base UE, 
the triangle DOU is equal to the triangle PUE, 
and the remaining angles are equal to the remaining angles ; 

[«■ 4] 
therefore the angle OUD is equal to the angle PUE. 















X>- 


^ 


\ 


P 




X 


^ 


^ 




^ 


1 


M 


\ 


L 


H 




v 


-A: 


\ 


\ 


R 


\ 


^S 


^ 


O 






For this reason DUE is a straight line. 0- 14] 

For the same reason, BSG is also a straight line, 
and BS is equal to SG. 

Now, since CA is equal and parallel to DB, 
while CA is also equal and parallel to EG, 
therefore DB is also equal and parallel to EG. [xi. 9] 



36a BOOK XI [xi. 38 

And the straight lines DE, BG join their extremities ; 

therefore DE is parallel to BG. [1. 33] 

Therefore the angle ED T is equal to the angle BG T, 
for they are alternate ; [1. 29] 

and the angle DTU is equal to the angle GTS. [1. 15] 

Therefore D TU, G TS are two I iangles which have two 
angles equal to two angles, and one side equal to one side, 
namely that subtending one of the equal angles, that is, D U 
equal to GS, 

for they are the halves of DE, BG ; 

therefore they will also have the remaining sides equal to the 

remaining sides. [1. 26] 

Therefore DT is equal to TG, and C/Tto TS. 

Therefore etc. 

Q. E. D. 

Euclid enunciates this proposition of a cube only, though it is true of any 
parallelepiped, no doubt because its truth for a cube is all that was wanted for 
the only proposition where it is needed, viz. xm. 17. 

Simson remarks that it should be proved that the straight lines bisecting 
the corresponding opposite sides of opposite planes art in one plane. This is, 
however, clear because e.g. since DK, CL are equal and parallel, KL is equal 
and parallel to CD. And, since KL, AB are both parallel to DC, KL is 
parallel to AB. And lastly, since KL, MNiXK both parallel to AB, KL is 
parallel to MNtsA therefore in one plane with it. 

The essential thing to be proved is that the plane passing through the 
opposite edges DB, EG passes through the straight line US, since, only if 
this be the case, can US, DG intersect one another. 

To prove this we have only to prove that, if DU, UE and BS, SG be 
joined, DUE and BSG are both straight lines. 

Now, since DO is parallel to PE, 

lDOU=lEPU. 

Thus, in the As DUO, EUP, two sides DO, OU are equal to two sides 
EP, PU, and the included angles are equal. 

Therefore &DUO ~ A EUP, 

DU^ UE, 
and i.nUO = LEUP, 

so that DUE is a straight line, bisected at U. Similarly BSG is a straight 
line, bisected at S. 

Thus the plane through DB, EG (DB, EG being equal and parallel) 
contains the straight lines DUE, BSG (which are therefore equal and parallel 
also) and also [xi. 7] the straight lines US, DG (which accordingly intersect). 

In As DTU, GTS, the angles UDT, SGT sue equal (being alternate), 
and the angles UTD, STG are also equal (being vertically opposite), while 
D U (half of DE) is equal to GS (half of BG). 



«• 38, 39] 



PROPOSITIONS 38, 39 



363 



Therefore [1. 16] the triangles D TU, GTS are equal in all respects, so that 
DT= TG, 
UT= TS. 

Proposition 39. 

If there be two prisms of equal height, and one have a 
parallelogram as base and the vtker a triangle, and if the 
parallelogram be double of the triangle, the prisms will be 
equal. 

Let ABCDEF, GHKLMN be two prisms of equal 
height, 

let one have the parallelogram AF as base, and the other the 
triangle GHK, 

and let the parallelogram AF be double of the triangle GHK; 
1 say that the prism ABCDEF is equal to the prism 
GHKLMN, 




For let the solids A0 t GP be completed. 

Since the parallelogram AF is double of the triangle GHK, 
while the parallelogram HK is also double of the triangle 
GHK, [,. 34] 

therefore the parallelogram AF is equal to the parallelogram 
HK 

But parallelepipedal solids which are on equal bases and 



of the same height are equal to one another ; 
therefore the solid AO is equal to the solid GP. 

And the prism ABCDEF is half of the solid A 0, 
and the prism GHKLMN is half of the solid GP ; 
therefore the prism ABCDEF is equal to the 
GHKLMN. 

Therefore etc. 

Q. E. 



[«• 3»] 



[xi. 28] 
prism 



364 BOOK XI [xi. 3 g 

This proposition is made use of in xn. 3, 4. The phraseology is interest- 
ins because we find one of the paralttlogrammic faces of one of the triangular 
prisms called its base, and the perpendicular on this plane from that vertex of 
either triangular face which is not in this plane the height. 

The proof is simple because we have only to complete parallelepipeds 
which are double the prisms respectively and then use xi. 31. It has to be 
borne in mind, however, that, if the parallelepipeds are not rectangular, the 
proof in xi. z8 is not sufficient to establish the fact that the parallelepipeds 
are double of the prisms, but has to be supplemented as shown in the note on 
that proposition. X11. 4 does, however, require the theorem in its general 
form. 






. 



BOOK XII. 



HISTORICAL NOTE. 

The predominant feature of Book xii. is the use of the method of 
exhaustion, which is applied in Propositions 2, 3 — 5, 10, 11, 12, and (in a 
slightly different form) in Propositions 16—18. We conclude therefore that 
for the content of this Book Euclid was greatly indebted to Eudoxus, to whom 
the discovery of the method of exhaustion is attributed. The evidence for 
this attribution comes mainly from Archimedes. (1) In the preface to On 
the Sphere and Cylinder 1., after stating the main results obtained by himself 
regarding the surface of a sphere or a segment thereof, and the volume and 
surface of a right cylinder with height equal to its diameter as compared with 
those of a sphere with the same diameter, Archimedes adds : " Having now 
discovered that the properties mentioned are true of these figures, I cannot 
feel any hesitation in setting them side by side both with my former investiga- 
tions and with those of the theorems of Eudoxus on solids which are held to be 
most irrefragably established, namely that any pyramid is one third part of the 
prism which has the same base with the pyramid and equal height [i.e. Eucl. 
xii. 7], and that any cone is one third part of the cylinder which has the same 
base with the tone and equal height [i.e. Eucl. xii. 10]. For, though these 
properties also were naturally inherent in the figures all along, yet they were 
in fact unknown to all the many able geometers who lived before Eudoxus 
and had not been observed by any one" (*) In the preface to the treatise 
known as the Quadrature of the Parabola Archimedes states the "lemma" 
assumed by him and known as the "Axiom of Archimedes" (see note on x. 1 
above) and proceeds : " Earlier geometers (oi irpoTtpov yeeu/« rpai) have also 
used this lemma; for it is by the use of this same lemma that they have 
shown that circles are to one another in the duplicate ratio of their diameters 
[Eucl. XII. 2], and that spheres are to one another in the triplicate ratio of their 
diameters [Eucl. xii. 18], and further that eitery pyramid is one third part of the 
prism which has the same base with the pyramid and equal height [Eucl. xii. 7]; 
also, that every cone is one third part of the cylinder which has the same base 
with the eone and equal height [Eucl, xii. 10] they proved by assuming a certain 
lemma similar to that aforesaid." Thus in the first passage two theorems of 
Eucl. xii. are definitely attributed to Eudoxus ; and, when Archimedes says, 
in the second passage, that " earlier geometers " proved these two theorems 
by means of the lemma known as the "Axiom of Archimedes" and of a 
lemma similar to it respectively, we can hardly suppose him to be alluding to 



366 BOOK. XII 

any other proof than that given by Eudoxus. As a matter of fact, the lemma 
used by Euclid to prove both propositions (xti. 3 — 5 and 7, and xn. 10) is the 
theorem of Eucl. x. 1. As regards the connexion between the two "lemmas" 
see note on x. 1. 

We are not, however, to suppose that none of the results obtained by 
the method of exhaustion had been discovered before the time of Eudoxus 
(fl. about 368 — s B.C.). Two at least are of earlier date, those of Eucl. xn. a 
and xn. 7. 

(a) Simplicius {Comment, in Aristot, Phys. p. 6i, ed. Diels) quotes 
Eudemus as saying, in his History of Geometry, that Hippocrates of Chios 
(fl. say 430 B.c.) first laid it down (Wno) that similar segments of circles are 
in the ratio of the squares on their bases and that he proved this (Ihtiwviv) by 
proving (U toO Scifat) that the squares on the diameters have the same ratio 
as the (whole) circles. We know nothing of the method by which Hippo- 
crates proved this proposition ; but, having regard to the evidence from 
Archimedes quoted above, it is not permissible to suppose that the method 
was the fully developed method of exhaustion as we know it. 

(b) As regards the two theorems about the volume of a pyramid and of 
a cone respectively, which Eudoxus was the first to prove, we now have 
authentic evidence in the short treatise by Archimedes discovered by Heiberg 
in a MS. at Constantinople in 1906 and published in Hermes the following 
year (see now Archimedis opera omnia, ed. Heiberg, 2, ed,, Vol. 11 ., 1913, 
pp. 425 — 507; '1". L. Heath, The Method of Archimedes, Cambridge, 191 2). 
The said treatise, complete in all essentials, bears the title 'Apxtw&ovs irtpt rmr 
fHflfin*** 8vapr)it£.Twv irpos "EpaToatiirrpr «#o&>y. This "Method" (or "Plan of 
attack "}, addressed to Eratosthenes, is none other than the ftfuSSio* on which, 
according to Suidas, Theodosius wrote a commentary, and which is several 
times cited by Heron in his Metrica ; its discovery adds a new and important 
chapter to the history of the integral calculus. In the preface to this work 
Archimedes alludes to the theorems which he first discovered by means of 
mechanical considerations, but proved afterwards by geometry, because the 
investigation by means of mechanics did not constitute a rigid proof; he 
observes, however, that the mechanical method is of great use for the discovery 
of theorems, and it is much easier to provide the rigid proof when the fact 
to be proved has once been discovered than it would be if nothing were 
known to begin with. He goes on : " Hence too, in the case of those 
theorems the proof of which was first discovered by Eudoxus, namely those 
relating to the cone and the pyramid, that the cone is one third part of the 
cylinder, and the pyramid one third part of the prism, having the same base 
and equal height, no small part of the credit will naturally be assigned to 
Democritus, who was the first to make the statement (of the fact) regarding 
the said figure [i.e. property], though without proving it." Hence the discovery 
of the two theorems must now be attributed to Democritus (fl. towards the 
end of 5th cent. B.C.). The words "without proving it " (x«P« £iroB«'£«i>5} do 
not mean that Democritus gave no sort of proof, but only that he did not give 
a proof on the rigorous lines required later ; for the same words are used by 
Archimedes of his own investigations by means of mechanics, which, however, 
do constitute a reasoned argument. The character of Archimedes' mechanical 
arguments combined with a passage of Plutarch about a particular question in 
infinitesimals said to have been raised by Democritus may perhaps give a clue 
to the line of Democritus' argument as regards the pyramid. The essential 



HISTORICAL NOTE 367 

feature of Archimedes' mechanical arguments in this tract is that he regards 
an area as the sum of an infinite number of straight lines parallel to one 
another and terminated by the boundary or boundaries of the closed figure 
the area of which is to be found, and a volume as the sum of an infinite 
number of plane sections parallel to one another : which is of course the same 
thing as taking (as we do in the integral calculus) the sum of an infinite 
number of strips of breadth dx (say), when dx becomes indefinitely small, or 
the sum of an infinite number of parallel laminae of depth dz (say), wlwn dz 
becomes indefinitely small. To give only one instance, we may take the 
case of the area of a segment of a parabola cut off by a chord. 

Let CBA be the parabolic segment, CE the tangent at C meeting the 







diameter EBD through the middle point of the chord CA in E, so that 

EB = BD. 

Draw AF parallel to ED meeting CE produced in F. Produce CB to 
H so that CK=.KH, where K is the point in which CH meets AF; and 
suppose CH 'to be a lever. 

Let any diameter MNPO be drawn meeting the curve in P and CF, CK, 
CA in M, N, respectively. 

Archimedes then observes that 

CA.AO = MO: OF 
(" for this is proved in a lemma "), 
whence 3. K : KJV= MO : OF, 

so that, if a straight line TG equal to PO be placed with its middle point at 
H, the straight line MO with centre of gravity at N, and the straight line TG 
■vith centre of gravity at H, will balance about K. 

Taking all other parts of diameters like PO intercepted between the curve 
and CA, and placing equal straight lines with their centres of gravity at H, 
these straight lines collected at H will balance (about K) all the lines like 
MO parallel to FA intercepted within the triangle CFA in the positions in 
which they severally lie in the figure. 

Hence Archimedes infers that an area equal to that of the parabolic 
segment hung at H will balance (about K) the triangle CFA hung at its 
centre of gravity, the point X (a point on CK such that CK=$XK), and 
therefore that 

(area of triangle CFA) : (area of segment) = HK: KX 

= 3:1. 



3<S8 BOOK XII 

from which it follows that 

area of parabolic segment = ^&ABC, 

The same sort of argument is used for solids, plane itetions taking the 
place of straight lines. 

Archimedes is careful to state once more that this method of argument 
does not constitute & proof. Thus, at the end of the above proposition about 
the parabolic segment, he adds : " This property is of course not proved by 
what has just been said; but it has furnished a sort of indication (1/i^acriV riva) 
that the conclusion is true." 

Let us now turn to the passage of Plutarch {De Comm. Not. adv. Stoieos 
xxx tx j) about Democritus above referred to. Plutarch speaks of Democritus 
as having raised the question in natural philosophy (^wo«5s) : " if a cone 
were cut by a plane parallel to the base [by which is clearly meant a plane 
indefinitely near to the base], what must we think of the surfaces of the 
sections, that they are equal or unequal ? For, if they are unequal, they will 
make the cone irregular, as having many indentations, like steps, and uneven- 
nesses ; but, if they are equal, the sections will be equal, and the cone will 
appear to have the property of the cylinder and to be made up of equal, not 
un equal circles, wh ic h is very absu rd , " The phrase" made up of eq uat . . . circl es " 
{i£ law o-iry *«/«»■<«,., mIkXcov) shows that Democritus already had the idea of 
a solid being the sum of an infinite number of parallel planes, or indefinitely 
thin laminae, indefinitely near together : a most important anticipation of the 
same thought which led to such fruitful results in Archimedes. If then one 
may hazard a conjecture as to Democritus' argument with regard to a pyramid, 
it seems probable that he would notice that, if two pyramids of the same 
height and equal triangular bases are respectively cut by planes parallel to the 
base and dividing the heights in the same ratio, the corresponding sections of 
the two pyramids are equal, whence he would infer that the pyramids are 
equal as being the sum of the same infinite number of equal plane sections 
or indefinitely thin laminae. (This would be a particular anticipation of 
Cavalieri's proposition that the areal or solid contents of two figures are equal 
if two sections of them taken at the same height, whatever the height may be, 
always give equal straight lines or equal surfaces respectively.) And 
Democritus would of course see that the three pyramids into which a prism 
on the same base and of equal height with the original pyramid is divided (as 
in End. xn. 7) satisfy this test of equality, so that the pyramid would be one 
third part of the prism. The extension to a pyramid with a polygonal base 
would be easy. And Democritus may have stated the proposition for the 
cone (of course without an absolute proof) as a natural inference from the 
result of increasing indefinitely the number of sides in a regular polygon 
forming the base of a pyramid. 









BOOK XII. PROPOSITIONS. 

Proposition i. 

Similar polygons inscribed in circles are to one another as 

the squares on the diameters. 

Let ABC, FGH be circles, 
let ABCDE, FGHKL be similar polygons inscribed in them, 
and let BM, GN be diameters of the circles ; 
I say that, as the square on BM is to the square on GN, so 
is the polygon ABCDE to the polygon FGHKL. 





For let BE, AM, GL, FN be joined. 

Now, since the polygon ABCDE is similar to the polygon 
FGHKL, 

the angle BAE is equal to the angle GFL, 
and, as BA is to AE, so is GF to FL. [vi. Def. i] 

Thus BAE, GFL are two triangles which have one angle 
equal to one angle, namely the angle BAE to the angle 
GFL, and the sides about the equal angles proportional ; 
therefore the triangle ABE is equiangular with the triangle 
FGL. [vi. 6} 

Therefore the angle AEB is equal to the angle FLG. 



370 BOOK XII [xii. i 

But the angle AEB is equal to the angle AMB, 
for they stand on the same circumference ; [ni. 27] 

and the angle FLG to the angle FNG ; 
therefore the angle AMB is also equal to the angle FNG. 

But the right angle BAM is also equal to the right angle 
GFN; [hi. 31] 

therefore the remaining angle is equal to the remaining angle. 

[«■ 3*] 

Therefore the triangle ABM is equiangular with the 
triangle FGN. 

Therefore, proportionally, as BM is to GN, so is BA 
to GF. [vi. 4] 

But the ratio of the square on BM to the square on GN 
is duplicate of the ratio of BM to GN, 

and the ratio of the polygon ABCDE to the polygon FGHKL 
is duplicate of the ratio of BA to GF; [vi. *o] 

therefore also, as the square on BM is to the square on GN, 
so is the polygon ABCDE to the polygon FGHKL. 

Therefore etc. 

Qt e. n. 

As, from this point onward, the text of each proposition usually occupies 
considerable space, I shall generally give in the notes a summary of [he 
argument, "to enable it to be followed more easily. 

Here we have to prove that a pair of corresponding sides are in the ratio 
of the corresponding diameters. 

Since i.s BAE, GFL are equal, and the sides about those angles 
proportional, 

As ABE, FGL are equiangular, 

so that ii AEB = l FLG 

Hence their equals in the same segments, l s AMB, FNG, are equal. 
And the right angles BAM, GFN ate equal- 
Therefore As ABM, FGN Me equiangular, so that 

BM : GN= BA : GF. 

The duplicates of these ratios are therefore equal, 
whence (polygon ABCDE) : (polygon FGHKL) 

= duplicate ratio of BA to GF 
- duplicate ratio of BM to GN 
= BM' : GN'. 



XII. t] 



PROPOSITIONS i, i 



37* 



Proposition 2. 
Circles are to one another as the squares on the diameters. 

Let ABCD, EFGH be circles, and BD, FH theii 
diameters ; 

I say that, as the circle ABCD is to the circle EFGH, so is 
the square on BD to the square on FH. 

A 




8 








T 

1 







For, if the square on BD is not to the square on FH as 
the circle ABCD is to the circle EFGH, 
then, as the square on BD is to the square on FH, so will 
the circle ABCD be either to some less area than the circle 
EFGH, or to a greater. 

First, let it be in that ratio to a less area 5. 

Let the square EFGH be inscribed in the circle EFGH \ 
then the inscribed square is greater than the half of the circle 
EFGH, inasmuch as, if through the points E, F, G, H we 
dn*w tangents to the circle, the square EFGH is half the 
square circumscribed about the circle, and the circle is less 
than the circumscribed square ; 

hence the inscribed square EFGH is greater than the half of 
the circle EFGH. 

Let the circumferences EF, EG, GH, HE be bisected at 
the points K, L, M, N, 

and let EK, KF, FL, LG, GM, MH, HN, NE be joined ; 
therefore each of the triangles EKF, FLG, GMH, HNE is 
also greater than the half of the segment of the circle about 
it, inasmuch as, if through the points K, L, M, N we draw 
tangents to the circle and complete the parallelograms on the 
straight lines EF, FG, GH, HE, each of the triangles EKF, 



373 BOOK XII [xn. 2 

FLG, GMH, HNE wilt be half of the parallelogram 
about it, 

while the segment about it is less than the parallelogram ; 
hence each of the triangles EKF, FLG, GMH, HNE 
is greater than the half of the segment of the circle 
about it. 

Thus, by bisecting the remaining circumferences and 
joining straight lines, and by doing this continually, we shall 
leave some segments of the circle which will be less than the 
excess by which the circle EFGH exceeds the area S. 

For it was proved in the first theorem of the tenth book 
that, if two unequal magnitudes be set out, and if from the 
greater there be subtracted a magnitude greater than the half, 
and from that which is left a greater than the half, and if this 
be done continually, there will be left some magnitude which 
will be less than the lesser magnitude set out. 

Let segments be left such as described, and let the 
segments of the circle EFGH on EK, KF, FL, LG, GM, 
MH, HN, NE be less than the excess by which the circle 
EFGH exceeds the area S. 

Therefore the remainder, the polygon EKFLGMHN, is 
greater than the area S. 

Let there be inscribed, also, in the circle A BCD the poly- 
gon AOBPCQDR similar to the polygon EKFLGMHN; 
therefore, as the square on BD is to the square on FH, so is 
the polygon AOBPCQDR to the polygon EKFLGMHN. 

[xn. i] 

But, as the square on BD is to the square on FH, so also 
is the circle ABCD to the area S ; 

therefore also, as the circle ABCD is to the area 5, so is the 
polygon AOBPCQDR to the polygon EKFLGMHN; 

[V.U] 

therefore, alternately, as the circle ABCD is to the polygon 
inscribed in it, so is the arei S to the polygon EKFLGMHN. 

[v. 16] 
But the circle ABCD is greater than the polygon inscribed 
in it; 

therefore the area 5 is also greater than the polygon 
EKFLGMHN. 



xii. 2] PROPOSITION 3 373 

But it is also less : 
which is impossible. 

Therefore, as the square on BD is to the square on FH, 
so is not the circle ABCD to any area less than the circle 
EFGH, 

Similarly we can prove that neither is the circle EFGH 
to any area less than the circle ABCD as the square on FH 
is to the square on BD. 

I say next that neither is the circle ABCD to any area 
greater than the circle EFGH as the square on BD is to the 
square on FH, 

For, if possible, let it be in that ratio to a greater area 5. 

Therefore, inversely, as the square on FH is to the square 
on DB, so is the area 5 to the circle ABCD. 

But, as the area S is to the circle ABCD, so is the circle 
EFGH to some area less than the circle ABCD ; 
therefore also, as the square on FH is to the square on BD, 
so is the circle EFGH to some area less than the circle 
ABCD: [v. ii] 

which was proved impossible. 

Therefore, as the square on BD is to the square on FH, 
so is not the circle ABCD to any area greater than the circle 
EFGH. 

And it was proved that neither is it in that ratio to any 
area less than the circle EFGH ; 

therefore, as the square on BD is to the square on FH, so ts 
the circle ABCD to the circle EFGH. 

Therefore etc. 

Q. E, D. 

Lemma. 

I say that, the area 5" being greater than the circle 
EFGH, as the area 5 is to the circle ABCD, so is the circle 
EFGH to some area less than tho circle ABCD, 

For let it be contrived that, as the area S is to the circle 
ABCD, so is the circle EFGH to the area T. 

I say that the area T is less than the circle ABCD. 

For since, as the area S is to the circle ABCD, so is the 
circle EFGH to the area T, 



374 BOOK XII [xii. 2 

therefore, alternately, as the area S is to the circle EFGH, so 
is the circle A BCD to the area T. [v. 16] 

But the area .S is greater than the circle EFGH ; 
therefore the circle ABCD is also greater than the area T. 

Hence, as the area S is to the circle ABCD, so is the 
circle EFGH to some area less than the circle ABCD. 

Q. E. D. 

Though this theorem is said to have been proved by Hippocrates, we may 
with tolerable certainty attribute the proof of it given by Euclid to Eudoxus, 
to whom xn. 7 For. and Xll. 10 (which Euclid proves in exactly the same 
manner} are specifically attributed by Archimedes. As regards the lemma 
used herein (Eucl. x. i) and the somewhat different lemma by means of which 
Archimedes says that the theorems of xii. 2, xll. 7 Par. and Xll. 18 were 
proved, see my note on x, 1 above. 

The first essential in this proposition is to prove that we can exhaust a 
circle, in the sense of x. 1, by successively inscribing in it regular polygons, 
each of which has twice as many sides as the preceding one. We take first 
an inscribed square, then bisect the arcs subtended by the sides and so form 
an equilateral polygon of eight sides, then do the same with the latter, forming 
a polygon of 16 sides, and so on. And we have to prove that what is left 
over when any one of these polygons is taken away from the circle is more 
than half exhausted when the next polygon is made and subtracted from the 
circle. 

Euclid proves that the inscribed square is greater than half the circle and 
that the regular octagon when subtracted takes away more than half of what 
was left by the square. He then infers that the same 
thing will happen whenever the number of sides is 
doubled. 

This can be seen generally by taking any arc of a 
circle cut off by a chord AB. Bisect the arc in C. 
Draw a tangent to the circle at C, and let AD, BE 
be drawn perpendicular to the tangent. Join AC, CB. 

Then D E is parallel to AB, since 
l ECB = l CAB, in alternate segment, [in. 32] 
= l CBA. [ill. 29, 1, 5] 

Thus A BED is a CD; 
and it is greater than the segment ACB, 

Therefore its half, the a A CB, is greater than half the segment. 

Thus, by x, 1, Euclid's construction of successive regular polygons in 
a circle, if continued far enough, will at length leave segments which are 
together less than any given area. 

Now let X, X' be the areas of the circles, d, d' their diameters, respectively. 

Then, if X : X' * d* : d' 1 , 

d*:d'* = X :S, 
where S is some area either greater or less than X '. 

I. Suppose 5 < X'. 

Continue the construction of polygons in X' until we arrive at one which 




xii. *] PROPOSITION * 375 

leaves over segments together less than the excess of X' over S, i.e. a polygon 
such that 

X ' > (polygon in X') > S. 

Inscribe in the circle X a polygon similar to that in X . 

Then (polygon in X) : (polygon in X') = d*:<f* [xn. 1] 

= X : S, by hypothesis ; 
and, alternately, 

(polygon in X ) : A" = (polygon in X') : S. 

But (polygon in X) < X; 

therefore (polygon in X') < S. 

But, by construction. (polygon in X) > S : 
which is impossible. 

Hence S cannot be less than X' as supposed. 

II. Suppose S> X\ 

Since d* : d* = X : S, 

we have, inversely, d' % • d* = S: X. 

Suppose that S:X=X':T, 

whence, since S>X', X > T. [v. 14] 

Consequently d'* : d* = X' : T, 

where T < X. 

This can be proved impossible in exactly the same way as shown in Part I. 

Hence S cannot be greater than X' as supposed. 

Since then .S' is neither greater nor less than X\ 

s^x; 

and therefore d> ; d' % = X : X'. 

With reference to the assumption that there is some space S such that 
d':d"^X: S t 
i.e. that there is a fourth proportional to the areas if, d' 1 , X, Simson observes 
that it is sufficient, in this and the like cases, that a thing made use of in the 
reasoning can possibly exist, though it cannot be exhibited by a geometrical 
construction. As regards the assumption see note on v. 18 above. 

There is grave reason for suspecting the genuineness of the Lemma at the 
end of the proposition ; though, if it be rejected, it will be necessary to delete 
the words "as was before proved " in corresponding places in xn. 5, 18. 

It will be observed that Euclid proves the impossibility in the second case 
by reducing it to the first If it is desired to prove the second case indepen- 
dently, we must circumscribe successive polygons to the circles instead ol 
inscribing them, in the way shown by Archimedes in his first proposition on 
the Measurement of a circle. Of course we require, as a preliminary, the 
proposition corresponding to xn. 1, that 
Similar polygons circumscribed about 
circles are to one another as the squares 
on the diameters. 

Let AB, A'ff be corresponding sides 
of the two similar polygons. Then L s 
OAB, OA'S are equal, since AO, A'O 
bisect equal angles. 




376 



BOOK XII 



[xii. 




Similarly lABO^lABO. 

Therefore As A OB, A' OB are similar, so that their areas are in the 
duplicate ratio of AB to A'B. 

The radii OC, OC drawn to the points of contact are perpendicular to 
AB, A'B, and it follows that 

AB : A'B = CO : CO. 

Thus the polygons are to one another in the duplicate ratio of the radii, 
and therefore of the diameters. 

Now suppose a square ABCD described about 
a circle. 

Make an octagon described about the circle by 
drawing tangents at the points £ etc., where OA etc. 
meet the circle. 

Then shall the tangent at £ cut off more than 
half of the area between AX', AH and the arc 
HEK. 

For the angle AEG is right, and is therefore 
? uEAG. 

Therefore AG>EG 

> G K. 

Therefore £.AGE> &EGK. 

Similarly &AFE > &EEH. 

Hence A A EG > £ (re-entrant quadrilateral A HEK), 

and a fortiori, A AEG > | (area between AH, AK sxA the arc). 

Thus the octagon takes from the square more than half the space between 
the square and the circle. 

Similarly, if a figure of 16 equal sides be circumscribed by cutting off 
symmetrically the corners of the octagon, it will take away more than half of 
the space between the octagon and circle. 

Suppose now, with the original notation, that 
P.d't^X-.S, 
where S is greater than X'. 

Continue the construction of circumscribed polygons about X' until the 
total area between the polygon and the circle is less than the difference 
between S and A", i.e. till 

S> (polygon about X') > X'. 

Circumscribe a similar polygon about X. 

Then (polygon about X ) : (polygon about X') = iP-.d' 1 

-X:S, by hypothesis, 
and, alternately, 

(polygon about X ) \ X = (polygon about X') : S. 

But (polygon about X) > X. 

Therefore (polygon about X') > S. 

But S > (polygon about X ') : [above] 

which is impossible. 

Hence S cannot be greater than X'. 



xti. a] 



PROPOSITION 2 



377 



Legend re proves this proposition by a method equally rigorous but not, I 
think, possessing any advantages over Euclid's. It depends on a lemma 
corresponding to Eucl. xu. 16, but with another part added to it. 

Initio concentric circles being given, we can always inscribe in the greater a 
regular polygon such that its sides do net meet the circumference of the lesser, and 
we can also circumscribe about the lesser a regular 
polygon such thai its sides do net meet the circum- u B 

ference of the greater. 

Let CA, CB be the radii of the circles. 

I. At A on the inner circle draw the tangent 
DE meeting the outer circle in D, E. 

Inscribe in the outer circle any of the regular 
polygons which we can inscribe, e.g. a square. 

Bisect the arc subtended by a side, bisect 
the half, bisect that again, and so on, until we 
arrive at an arc less than the arc DBE. 

Let this arc be MN, and suppose it so placed 
that B is its middle point. 

Then the chord MN is clearly more distant from the centre C than DE 
is ; and the regular polygon, of which MN is a side, does not anywhere meet 
the circumference of the inner circle. 

II. Join CM, CN, meeting DE in P, Q. 

Then PQ will be the side of a polygon circumscribed about the inner 
circle and similar to the polygon inscribed in the outer ; 
and the circumscribed polygon of which PQ is a side will not anywhere meet 
the outer circle. 

Legendre now proves xil. a after the following manner. 
For brevity, let us denote the area of the circle with radius CA by 
(circ CA), 

Then it is required to prove that, if OB be the radius of a second circle, 
(circ. CA) : (circ OB) = CA* : OB 1 . 









Suppose, if possible, that this relation is not true. Then CA 1 will be to 
OB 1 as (circ, CA) is to an area greater or less than (circ, OB). 

I. Suppose, first, that 

CA' t OB 1 = (circ CA) ■ (circ OB), 
where OD is less than OB. 



378 BOOK XII [SO. 2, 3 

Inscribe in the circle with radius OS a regular polygon such that its sides 
do not anywhere meet the circumference of the circle with centre OD ; 

[Lemma] 
and inscribe a similar polygon in the other circle. 

The areas of the polygons will then be in the duplicate ratio of CA to OB, 
or [xil. 1] 

{polygon in circ. CA) : (polygon in circ. OS) 
= CA?.OB* 

= (circ. CA) : (circ. OD), by hypothesis. 
But this is impossible, because the polygon in (circ. CA) is less than (cite. 
CA), but the polygon in (circ. OB) is greater than (circ. OD). 

Therefore CA 1 cannot be to OS 1 as (circ. CA) ts to a less circle than 
(circ. OS). 

1 1. Suppose, if possible, that 

CA* : OS 1 = (circ. CA) : (some circle > circ. OS). 
Then inversely 

OS* : CA* - (circ. OS) : (some circle < circ. CA), 
and this is proved impossible exactly as in Part I. 

Therefore CA* : OS* = (circ. CA) : (circ. OS). 



PROI'OSITION 3. 

Any pyramid which has a triangular base is divided into 
two pyramids equal and similar to one another, similar to the 
whole and having triangular bases, and into two equal prisms ; 
and the two prisms are greater than the half of the whole 
pyramid. 

Let there be a pyramid of which the triangle ABC is the 
base and the point D the vertex ; 
I say that the pyramid ABCD is 
divided into two pyramids equal to 
one another, having triangular bases 
and similar to the whole pyramid, 
and into two equal prisms ; and the 
two prisms are greater than the half 
of the whole pyramid. 

For let AB, BC, CA, AD, DB, 
DC be bisected at the points E, F, 
G, H, K, L, and let HE, EG, GH, HK, KL, LH, KF, FG 
be joined. 

Since AE is equal to EB, and AH to DH, 
therefore EH is parallel to DB. [n 3] 




xii. 3] PROPOSITIONS 2, 3 379 

For the same reason 
HK is also parallel to AB, 

Therefore HEBK is a parallelogram ; 
therefore HK is equal to EB. [t- 34] 

But EB is equal to EA ; 
therefore AE is also equal to HK. 

But AH is also equal to HD ; 
therefore the two sides EA, AH are equal to the two sides 
KH, HD respectively , 

and the angle EAH is equal to the angle KHD ; 
therefore the base EH is equal to the base KD. [t 4] 

Therefore the triangle A EH is equal and similar to the 
triangle HKD. 

For the same reason 
the triangle AHG is also equal and similar to the triangle 
HLD. 

Now, since two straight lines EH, HG meeting one 
another are parallel to two straight lines KD, DL meeting 
one another, and are not in the same plane, they w:il contain 
equal angles. [xi. 10] 

Therefore the angle EHG is equal to the angle KDL. 

And, since the two straight lines EH, HG are equal to the 
two KD, DL respectively, 
and the angle EHG is equal to the angle KDL, 
therefore the base EG is equal to the base KL ; [1. 4] 

therefore the triangle EHG is equal and similar to the 
triangle KDL. 

For the same reason 

the triangle AEG is also equal and similar to the triangle 
HKL. 

Therefore the pyramid of which the triangie AEG is the 
base and the point H the vertex is equal and similar to the 
pyramid of which the triangle HKL is the base and the point 
D the vertex. [xi. Def. 10] 

And, since HK has been drawn parallel, to AB, one of the 
sides of the triangle ADB, 



380 BOOK XII [xn. 3 

the triangle ADB is equiangular to the triangle DHK, [i. 29] 

and they have their sides proportional ; 

therefore the triangle ADB is similar to the triangle DHK. 

[vi. Del. 1] 

For the same reason 
the triangle DBC is also similar to the triangle DKL, and 
the triangle ADC to the triangle DLH. 

Now, since the two straight lines BA, AC meeting one 
another are parallel to the two straight lines KH, HL meeting 
one another, not in the same plane, they will contain equal 
angles. [xi. 10] 

Therefore the angle BAC is equal to the angle KHL. 

And, as BA is to AC, so is KH to HL ; 
therefore the triangle ABC is similar to the triangle HKL. 

Therefore also the pyramid of which the triangle ABC is 
the base and the point D the vertex is similar to the pyramid 
of which the triangle HKL is the base and the point D the 
vertex. 

But the pyramid of which the triangle HKL is the base 
and the point D the vertex was proved similar to the pyramid 
of which the triangle AEG is the base and the point H the 
vertex. 

Therefore each of the pyramids AEGH, HKLD is 
similar to the whole pyramid A BCD. 

Next, since BE is equal to EC, 
the parallelogram EBFG is double of the triangle GEC. 

And since, if there be two prisms of equal height, and one 
have a parallelogram as base, and the other a triangle, and if 
the parallelogram be double of the triangle, the prisms are 
equal, [xi. 39] 

therefore the prism contained by the two triangles BKF, 
EHG, and the three parallelograms EBFG, EBKH, HKEG 
is equal to the prism contained by the two triangles GEC, 
HKL and the three parallelograms KECL, LCGH, HKEG. 
And it is manifest that each of the prisms, namely that in 
which the parallelogram EBFG is the base and the straight 
line HK is its opposite, and that in which the triangle GEC is 
the base and the triangle HKL its opposite, is greater than 
each of the pyramids of which the triangles AEG, HKL are 
the bases and the points H, D the vertices, 



xii. 3] PROPOSITION 3 381 

inasmuch as, if we join the straight lines £F, EK, the prism 
in which the parallelogram EBFG is the base and the straight 
line HKits opposite is greater than the pyramid of which the 
triangle EBF is the base and the point K the vertex. 

But the pyramid of which the triangle EBF is the base 
and the point K the vertex is equal to the pyramid of which 
the triangle AEG is the base and the point H the vertex ; 
for they are contained by equal and similar planes. 

Hence also the prism in which the parallelogram EBFG 
is the base and the straight line HK its opposite is greater 
than the pyramid of which the triangle AEG is the base and 
the point H the vertex. 

Bat the prism in which the parallelogram EBFG is the 
base and the straight line HK its opposite is equal to the 
prism in which the triangle GFC is the base and the triangle 
HKL its opposite, 

and the pyramid of which the triangle AEG is the base and 
the point H the vertex is equal to the pyramid of which the 
triangle HKL is the base and the point D the vertex. 

Therefore the said two prisms are greater than the said 
two pyramids of which the triangles AEG, HKL are the 
bases and the points H, D the vertices. 

Therefore the whole pyramid, of which the triangle ABC 
is the base and the point D the vertex, has been divided into 
two pyramids equal to one another and into two equal prisms, 
and the two prisms are greater than the half of the whole 
pyramid. 

Q. e. d. 

We will denote a pyramid with vertex D and base ABC by D (ABC) or 
D-ABC and the triangular prism with triangles GCF, HLK for bases by 
(GCF, HLK). 

The following are the steps of the proof. 

I. To prove pyramid H(AEG) equal and similar to pyramid D(HKL). 
Since sides of &.DAB are bisected at N, E, K, 
HE || DB, and HK\\ AB. 
Hence HK=EB = EA, 

HE = KB = DK. 
Therefore (1) As HAE, DHKz.it equal and similar. 
Similarly (2) As HAG, DHL are equal and similar. 
Again, Lff, HK are respectively || to GA, AE in a different plane ; 
therefore l. GAE = L LHK. 



382 BOOK XII [xil. 3, 4 

And LH, If/Cue respectively equal to GA, AE. 
Therefore (3) As GAE, LHK are equal and similar. 
Similarly (4) As HGE, £>LK axe equal and similar. 
Therefore [xi. Def. 10] the pyramids //(AEG) and D(HKL) are equal 
and similar. 

II. To prove the pyramid D (HKL) similar to the pyramid D (ABC). 
(1) The As DHK, DAB are equiangular and therefore similar. 
Similarly {*) As DLH, DCA are similar, as also (3) the As DLK, DCB. 
Again, BA, AC are respectively parallel to KH, HL in a different plane ; 

therefore lBAC^l. KHL. 

And BA : AC= KH : HL. 

Therefore {4) As BAC, KHL are similar. 

Consequently the pyramid D(ABC) is similar to the pyramid D (HKL), 
and therefore also to the pyramid H(AEG). 

III. To prove prism (GCF, HLK) equal to prism {HGE, KFB). 

The prisms may be regarded as having the same hd^ht (the distance 
between the planes HKL, ABC) and having for bases (i) the ACC^and 
(2) the O EBFG, which is the double of the A CGF. 

Therefore, by XI. 39, the prisms are equal. 

IV. To prove the prisms greater than the small pyramids. 

Prism (HGE, KFB) is clearly greater than pyramid K(EFB) and there- 
fore greater than pyramid H(AEG). 

Therefore each of the prisms is greater than each of the small pyramids ; 
and the sum of the two prisms is greater than the sum of the two small 
pyramids, which, with the two prisms, make up the whole pyramid. 

Proposition 4. 

If there be two pyramids of the same height which have 
triangular bases, and each of them be divided into two pyramids 
equal to one another and similar to the whole, and into two 
equal prisms, then, as the base of the one pyramid is to the 
base of the other pyramid, so will all the prisms in the one 
pyramid be to all the prisms, being equal in multitude, in the 
other pyramid. 

Let there be two pyramids of the same height which 
have the triangular bases ABC, DEF, and vertices the 
points G, H, 

and let each of them be divided into two pyramids equal to 
one another and similar to the whole and into two equal 
prisms ; [xil 3] 

I say that, as the base ABC is to the base DEF, so are 
all the prisms in the pyramid ABCG to all the prisms, being 
equal in multitude, in the pyramid DEFH, 



Xii. 4 J 



PROPOSITIONS 3, 4 



For, since BO is equal to OC, and AL to LC, 
therefore LO is parallel to AB, 
and the triangle ABC is similar to the triangle LOC. 



3»3 








For the same reason 
the triangle DEFis also similar to the triangle RVF. 

And, since BC is double of CO, and EF of FV, 
therefore, as BC is to CO, so is EF to i 7 ^. 

And on BC, CO are described the similar and similarly 
situated rectilineal figures ABC, LOC, 

and on EF, FV the similar and similarly situated figures 
DEF, RVF; 

therefore, as the triangle ABC is to the triangle LOC, so is 
the triangle DEFto the triangle RVF; [vi. 22] 

therefore, alternately, as the triangle ABC is to the triangle 
DEF, so is the triangle LOC to the triangle RVF. [v. 16] 

But, as the triangle LOC is to the triangle RVF, so Is 
the prism in which the triangle LOC is the base and PMN its 
opposite to the prism in which the triangle R VF is the base 
and STU its opposite ; [Lemma following] 

therefore also, as the triangle ABC is to the triangle DEF, 
so is the prism in which the triangle LOC is the base and 
PMN its opposite to the prism in which the triangle RVF 
is the base and S TU its opposite. 

But, as the said prisms are to one another, so is the prism 
in which the parallelogram KBOL is the base and the straight 
line PM its opposite to the prism in which the parallelogram 
QEVR is the base and the straight line ST its opposite. 

[xi. 39 ; cf. xii. 3] 



384 BOOK XII [xii. 4 

Therefore also the two prisms, that in which the parallelo- 
gram KBOL is the base and PM its opposite, and that in 
which the triangle LOC is the base and PMN its opposite, 
are to the prisms in which QEVR is the base and the straight 
line ST its opposite and in which the triangle RVF is the 
base and STU its opposite in the same ratio [v. u] 

Therefore also, as the base ABC is to the base DEF, so 
are the said two prisms to the said two prisms. 

And similarly, if the pyramids PMNG, STUH be divided 
into two prisms and two pyramids, 

as the base PMN is to the base STU, so will the two prisms 
in the pyramid PMNG be to the two prisms in the pyramid 
STUH. 

But, as the base PMN is to the base STU, so is the base 
ABC to the base DEF ; 

for the triangles PMN, STU are equal to the triangles LOC, 
R VF respectively. 

Therefore also, as the base ABC is to the base DEF, so 
are the four prisms to the four prisms. 

And similarly also, if we divide the remaining pyramids 
into two pyramids and into two prisms, then, as the base 
ABC is to the base DEF, so will all the prisms in the 
pyramid ABCG be to all the prisms, being equal in multitude, 
in the pyramid DEFH. 

Q. E. D. 

Lemma. 

But that, as the triangle LOC is to the triangle RVF, 
so is the prism in which the triangle LOC is the base and 
PMN its opposite to the prism in which the triangle RVF is 
the base and STU its opposite, we must prove as follows. 

For in the same figure let perpendiculars be conceived 
drawn from G, H to the planes ABC, DEF; these are of 
course equal because, by hypothesis, the pyramids are of equal 
height. 

Now, since the two straight lines <7Cand the perpendicular 
from G are cut by the parallel planes ABC, PMN, 
they will be cut in the same ratios. [xt. 17) 



xii. 4] PROPOSITION 4 385 

And GC is bisected by the plane PMN at N \ 

therefore the perpendicular from G to the plane ABC will 
also be bisected by the plane PMN. 

For the name reason 

the perpendicular from H to the plane DEF will also be 
bisected by the plane ST if. 

And the perpendiculars from G t H to the planes ABC, 
DEF are equal ; 

therefore the perpendiculars from the triangles PMN, STU 
to the planes ABC, DEF are also equal. 

Therefore the prisms in which the triangles LOC, RVF 
are bases, and PMN, STU their opposites, are of equal 
height. 

Hence also the parallelepi pedal solids described from the 
said prisms are of equal height and are to one another as their 
bases ; [xi. 3*] 

therefore their halves, namely the said prisms, are to one another 
as the base LOC is to the base RVF, 

Q. E. D. 

We can incorporate the lemma at the end of the proposition and sum- 
marise the proof thus. 

Since LO is parallel to AB, 

As ABC, LOC ate simitar. 
In like manner As DEF, RVF are similar. 

And, since BC : CO = EF : FV, 

A ABC : A LOC = A DEF: A RVF, [vi. 22] 

and, alternately, 

A ABC : &DEF= ALOC.ARVF. 
Now the prisms {LOC, PUN) and {RVF, STU) are equal in height : 
for the perpendiculars from G, H on the bases ABC, DEF are divided by 
the planes PMN, STU (parallel to the bases) in the same proportion as GC, 
fi/Faxe divided by those planes [xi. 17], i.e. they are bisected j 
hence the heights of the prisms, being half the equal heights of the pyramids, 
are equal. 

And the prisms are the halves respectively of parallelepipeds of the same 
height on parallelogram mic bases double of the As LOC, RVF respectively : 

[xi. 28 and note] 
hence they are in the same ratio as those parallelepipeds, and therefore as 
their bases [xi. J»J 
Therefore 

(prism LOC, PMN) : (prism RVF, STU) = A LOC: A RVF 

= AABC:ADEF. 



386 



BOOK XII 



[xn. 4. 5 



And since the other prisms in the pyramids are equal to these prisms 
respectively, 

(sum of prisms in GABC) : (sum of prisms in HDEF) =AABC:&£)EF. 
Similarly, if the pyramids GPMN, HSTU be divided in like manner, and 
also the pyramids PAKL, SDQJt, we shall have e.g. 

(sum of prisms in GPMN) : (sum of prisms in HSTU) = &PMN:h.STU 

= AABC:A£>EF, 
and similarly for the second pair of pyramids. 

The process may be continued indefinitely, and we shall always have 
(sum of prisms in GABC) : (sum of prisms in HDEF) =A ABC :ADEF. 



Proposition 5. 



Pyramids •which are of the same height and have triangular 
bases are to one another as the bases. 

Let there be pyramids of the same height, of which the 
triangles ABC, DEF are the bases and the points G, H the 
vertices ; 

I say that, as the base ABC ts to the base DEF, so is the 
pyramid ABCG to the pyramid DEFH, 






For, if the pyramid ABCG is not to the pyramid DEFH 
as the base ABC is to the base DEF, 
then, as the base ABC is to the base DEF, so will the 
pyramid ABCG be either to some solid less than the pyramid 
DEFH or to a greater. 

Let it, first, be in that ratio to a less solid W, and let the 
pyramid DEFH be divided into two pyramids equal to one 
another and similar to the whole and into two equal prisms ; 

then the two prisms are greater than the half of the whole 
pyramid, [xii. 3] 



xn. 5] PROPOSITIONS 4, s 387 

Again, let the pyramids arising from the division be 
similarly divided, 

and let this be done continually until there are left over from 
the pyramid DEFH some pyramids which are less than the 
excess by which the pyramid DEFH exceeds the solid W. 

[x, 1] 

Let such be left, and let them be, for the sake of argument, 
DQRS, STUH\ 

therefore the remainders, the prisms in the pyramid DEFH, 
are greater than the solid IV. 

Let the pyramid ABCG also be divided similarly, and a 
similar number of times, with the pyramid DEFH; 
therefore, as the base ABC is to the base DEF, so are the 
prisms in the pyramid ABCG to the prisms in the pyramid 
DEFH. [xii. 4] 

But, as the base ABC is to the base DEF, so also is the 
pyramid ABCG to the solid W\ 

therefore also, as the pyramid ABCG is to the solid W, so 
are the prisms in the pyramid ABCG to the prisms in the 
pyramid DEFH; [v. 11] 

therefore, alternately, as the pyramid ABCG is to the prisms 
in it, so is the solid W to the prisms in the pyramid DEFH. 

[v. .6] 

But the pyramid ABCG is greater than the prisms in it ; 
therefore the solid W is also greater than the prisms in the 
pyramid DEFH. 

But it is also less : 
which is impossible. 

Therefore the prism ABCG is not to any solid less than 
the pyramid DEFH as the base ABC is to the base DEF. 

Similarly it can be proved that neither is the pyramid 
DEFH to any solid less than the pyramid ABCG as the base 
DEF is to the base ABC. 

I say next that neither is the pyramid ABCG to any 
solid greater than the pyramid DEFH as the base ABC is 
to the base DEF. 

For, if possible, let it be in that ratio to a greater solid W; 
therefore, inversely, as the base DEF is to the base ABC, 
so is the solid W to the pyramid ABCG. 



3 88 BOOK XII [xii. s 

But, as the solid W is to the solid ABCG, so is the 
pyramid DEFH to some solid less than the pyramid ABCG, 
as was before proved ; [xn. a, Lemma] 

therefore also, as the base DEF is to the base ABC, so is 
the pyramid DEFH to some solid less than the pyramid 
ABCG: [v. m] 

which was proved absurd. 

Therefore the pyramid ABCG is not to any solid greater 
than the pyramid DEFH as the base ABC is to the base 
DEF. 

But it was proved that neither is it in that ratio to a less 
solid. 

Therefore, as the base ABC is to the base DEF, so is 
the pyramid ABCG to the pyramid DEFH 

Q. E. D. 

In the two preceding propositions it has been shown how we can divide a 
py ram id with a triangular base into (1) two equal prisms which are together 
greater than half the pyramid and (2) two equal pyramids similar to the 
original one, and that, if this process be continued with the two pyramids, 
then with the four resulting pyramids, and so on, and if, further, another 
pyramid of the same height as the original one be similarly divided, the sub- 
division being made the same number of times, the sum of all the prisms in 
one pyramid is to the sum of all the prisms in the other as the base of the 
first is to the base of the second. 

We can now prove in the manner of xn. 2 that the volumes of the 
pyramids themselves are as the bases. 

Let us call the pyramids P, P and their respective bases B, B. 

If P.P + B.&, 

suppose that B-.B-P; W. 

I. Let W be < P. 

Divide P into two prisms and two pyramids, subdivide the latter similarly, 
and so on, until the sum of the pyramids remaining is less than the difference 
between P and W[x, 1], so that 

P > (prisms in P) a- W. 

Then divide P similarly, the same number of times. 

Now (prisms in P) : (prisms in P) = B : B [xn. 4] 

= P : W, by hypothesis, 
and, alternately, 

(prisms in P):P= (prisms in P) : W. 

But (prisms in P) < P; 

therefore (prisms in P) < W. 

But, by construction, (prisms in P) > W. 

Hence W cannot be less than P. 



xii. s] 



PROPOSITION s 



3»9 



II. Suppose, if possible, that W> f. 

Then, inversely, B : B = W.P, 

= P": V t 
where V is some solid less than P. [Cf. xii. z. Lemma, and note.] 

But this can be proved impossible exactly as in Part I. 

Therefore Wis neither less nor greater than P", 

sothat B.B^P.P. 

Legend re, followed by the American editors already mentioned, and by 
others, approaches the subject by a different route, proving the following 
propositions. 

r. If a pyramid be cut by a plant parallel to the base, (a) the lateral edges 
and the height will be cut in the same proportion, {b) the section by the plant 
will be a polygon similar to the base. 




(a) Since a lateral face VAB of the pyramid V(ABCDE) is cut by two 
parallel planes in AB, ab, 

AB\\ab; 
Similarly BC j be, and so on. 
Therefore VA : Va = VB: Vb= VC: Vem„„ 

And, if VO the height be cut in O, 0, 

BO I! bo j and each of the above ratios is equal to VO : Vo. 

(b) Since BA \\ba, znd BC\\ be, 

(,ABC= Lobe. [xi. 10] 

Similarly for all the other angles of the polygons, which are therefore 
equiangular. 

Also, by similar triangles, 

VA : Va = AB : ab, 
and so on. 

Therefore, by the ratios above, 

AB:ab = BC; bc= .... 
Therefore the polygons are similar. 

2. If two pyramids of the same height be cut by plants which art at tht 
same perpendicular distance from the vertices, the sections are as the rcsptctivt 
baits. 



39 o 



BOOK XII 



[xn. 5 



For, if we place the pyramids so that the vertices coincide and the bases 
are in one plane, the planes of the sections will coincide. 

If, e.g. t the base of the second pyramid be XYZ and the section xyt, we 
shall have, by the argument of the last proposition, 

VX : Vx = VY : Vy = VZ: V* = VO : Ve= VA : Va = ... , 
and X YZ, xyz will be similar. 

Now (polygon ABCDE) : (polygon abate) = AS 1 : aP 

= FA' : Va 1 , 
and A XYZ \ bxy* = XY* : xy* 

= VX* : Vx? 
= VA* : Va*. 
Therefore 

(polygon ABCDE) : (polygon abcde) =AXYZ:& xyt. 
As a particular case, if the bases of the two pyramids are equivalent, the 
sections art aha equivalent. 

3. 7itx> triangular pyramids which have equivalent bases and equal heights 
are equivalent. 

Let VABC, vabc be pyramids with equivalent bases ABC, abe, which for 
convenience we will suppose placed in one plane, and let TA be the common 

height. 





Then, if the pyramids are not equivalent, one must be greater than the other. 

Let VABC be the greater ; and let AX be the height of a prism on ABC 
as base which is equal in volume to the difference of the pyramids. 

Divide the height AT into equal parts such that each is less than AX, and 
let each part be equal to z. 

Through the points of division draw planes parallel to the bases cutting 
both pyramids in the sections DEF, GHI,... and def, ghi, .... 

The sections DEF, def mi\\ then be equivalent ; so will the sections GHI, 
ghi, and so on. [(2) above] 

On the triangles ABC, DEF, GHI, ... as bases draw exterior prisms 
having for edges the parts AD, DG, GIC, ... of the edge A V; 



xii. jj PROPOSITION 5 391 

and on the triangles def, ghi, ... as bases draw interior prisms having for edges 
the parts aif, dg, ... of av. 

All the partial prisms will then have the same height 1. 

Now the sum of the exterior prisms of the pyramid VABC is greater than 
that pyramid ; 

and the sum of the interior prisms in the pyramid vabe is less than that 
pyramid. 

Consequently the difference between the sum of the first set of prisms and 
the sum of the second set of prisms is greater than the difference between the 
two pyramids. 

Again, if we start from the bases ABC, air, the second exterior prism 
DEJFG is equivalent to the first interior prism defa, since their bases are 
equivalent and they have the same height z. [xt. 28 and note ; xi. 32] 

Similarly the third exterior prism is equivalent to the second interior 
prism, and so on, until we arrive at the last of each. 

Therefore the prism ABCD % the first exterior prism, is the difference 
between the sums of the exterior and interior prisms respectively. 

Therefore the difference between the two pyramids is less than the prism 
ABCD, which should therefore be greater than the prism with base ABC 
and height AX. 

But the prism ABCD is, by hypothesis, less than the latter prism : 
which is impossible. 

Consequently the pyramid VABC cannot be greater than the pyramid 
k. 

Similarly it may be proved that vabe cannot be greater than VABC. 

Therefore the pyramids are equivalent. 

Legendre next establishes a proposition corresponding to Eucl. XII. 7, vu. 

4. Any triangular pyramid is one third of the triangular prism on the same 
base and of the same height, 

and from this he deduces that 

Cor. The volume of a triangular pyramid is equal to a third of the product 
of its base by its height. 

He has previously proved that the volume of a triangular prism is equal to 
the product of its base and height, since {1) the prism is half of a parallele- 
piped of the same height and with a parallelogram for base which is double of 
the base of the prism, and {2) this parallelepiped can be transformed into an 
equivalent rectangular parallelepiped with the same height and an equivalent 
base. 

The theorem (4} is then extended to any pyramid in the proposition 

5. Any pyramid has for its measure the third part of the product of its base 
and its height, from which follow 

Cor. I. Any pyramid is the third part of the prism on the same base and 
of the same height. 

Cor. II. Two pyramids of the same height are to one another as their 
bases, and huo pyramids on the same base are to one another as their heights. 

The first part of the second corollary corresponds to the present 
proposition as extended by the next, xu. 6. 



39* BOOK XII [xn, 6 

Proposition 6. 

Pyramids which are of the same height and have polygonal 
bases are to one another as the bases. 

Let there be pyramids of the same height of which the 
polygons ABCDE, FGHKL are the bases and the points 
M, N the vertices ; 

I say that, as the base ABCDE is to the base FGHKL, 
so is the pyramid ABCDEM to the pyramid FGHKLN. 







For let AC, AD, FH, FK be joined. 

Since then ABCM, ACDM are two pyramids which have 
triangular bases and equal height, 

they are to one another as the bases ; [xn. 5] 

therefore, as the base ABC is to the base A CD, so is the 
pyramid ABCM to the pyramid ACDM. 

And, componendo, as the base A BCD is to the base A CD, 
so is the pyramid ABCDM to the pyramid A CDM. [v. 18] 

But also, as the base A CD is to the base ADE, so is the 
pyramid ACDM to the pyramid ADEM. [xn. 5] 

Therefore, ex aequali, as the base A BCD is to the base 
ADE, so is the pyramid ABCDM to the pyramid ADEM. 

[v. m] 

And again componendo, as the base ABCDE is to the 
base ADE, so is the pyramid ABCDEM to the pyramid 
ADEM. [v. 18} 

Similarly also it can be proved that, as the base FGHKL 
is to the base FGH, so is the pyramid FGHKLN to the 
pyramid FGHN. 



xii. 6] PROPOSITION 6 393 

And, since A DEM, FGHN are two pyramids which have 
triangular bases and equal height, 

therefore, as the base ADE is to the base FGH, so is the 
pyramid A DEM to the pyramid FGHN. [xii. 5] 

But, as the base ADE is to the base ABCDE, so was 
the pyramid ADEM to the pyramid ABCDEM. 

Therefore also, ex aequali, as the base ABCDE is to the 
base FGH, so is the pyramid ABCDEM to the pyramid 
FGHN. [v. i»] 

But further, as the base FGH is to the base FGHKL, so 
also was the pyramid FGHN to the pyramid FGHKL N. 

Therefore also, ex atquali, as the base ABCDE is to the 
base FGHKL, so is the pyramid ABCDEM to the pyramid 
FGHKLN. [v. »] 

Q. E. D. 

It will be seen that, in order to obtain the proportion 

(base ABCDE) : A ADE - (pyramid M ABCDE) : (pyramid MADE), 

Euclid employs v. 18 (componendo\ twice over, with an ex atquali step [v. aj] 
intervening. 

We might arrive at it more concisely by using v. 24 extended to any 
number of antecedents, 

Thus 

AABC-.AADE = (pyramid MABC) : (pyramid MADE), 

AACD :AADE = (pyramid MA CD) : (pyramid MADE), 

and lastly 

A ADE : AADE = (pyramid MADE) ; (pyramid MADE). 

Therefore, adding the antecedents [v, 24], we have 

(polygon ABCDE) :AADE = (pyramid M ABCDE) : (pyramid MADE). 

Again, since the pyramids MADE, NFGHare of the same height, 

A ADE ; AFGH^ (pyramid MADE) ; (pyramid JVFGH). 

Lastly, using the same argument for the pyramid NFGHKL as for 
MABCDE, and inverting, we have 

A FGH: (polygon FGHKL) m (pyramid NFGH) : (pyramid NFGHKL). 

Thus from the three proportions, ex aequali, 

(polygon ABCDE) : (polygon FGHKL) 

m (pyramid MABCDE) : (pyramid JVFGHJpL). 



394 BOOK XII [xn. 7 







Proposition 7. 

Any prism which has a triangular base is divided into three 
pyramids equal to one another which have triangular bases. 

Let there be a prism in which the triangle ABC is the 
base and DEF its opposite ; 
I say that the prism ABCDEF is 
divided into three pyramids equal to 
one another, which have triangular 
bases. 

For let BD, EC, CD be joined. 

Since ABED is a parallelogram, 
and BD is its diameter, 
therefore the triangle ABD is equal 
to the triangle EBD \ [1. 34] 

therefore also the pyramid of which the triangle ABD is the 
base and the point C the vertex is equal to the pyramid of 
which the triangle DEB is the base and the point C the 
vertex. [so. 5] 

But the pyramid of which the triangle DEB is the base 
and the point C the vertex is the same with the pyramid of 
which the triangle EBC is the base and the point D the 
vertex ; 
for they are contained by the same planes. 

Therefore the pyramid of which the triangle ABD is the 
base and the point C the vertex is also equal to the pyramid 
of which the triangle EBC is the base and the point D the 
vertex. 

Again, since FCBE is a parallelogram, 
and CE is its diameter, 
the triangle CEF is equal to the triangle CBE. [1. 34] 

Therefore also the pyramid of which the triangle BCE is 
the base and the point D the vertex is equal to the pyramid 
of which the triangle ECF is the base and the point D the 
vertex. [*«, 5] 

But the pyramid of which the triangle BCE is the base 
and the point D the vertex was proved equal to the pyramid 
of which the triangle ABD is the base and the point C the 
vertex ; 



xn. r, 8] PROPOSITIONS 7, 8 39S 

therefore also the pyramid of which the triangle CEF is the 
base and the point D the vertex is equal to the pyramid of 
which the triangle ABD is the base and the point C the 
vertex ; 

therefore the prism ABCDEF has been divided into three 
pyramids equal to one another which have triangular bases. 

And, since the pyramid of which the triangle ABD is the 
base and the point C the vertex is the same with the pyramid 
of which the triangle CAB is the base and the point D the 
vertex, 

for they are contained by the same planes, 
while the pyramid of which the triangle ABD is the base and 
the point C the vertex was proved to be a third of the prism 
in which the triangle ABC is the base and DBF its opposite, 
therefore also the pyramid of which the triangle ABC is the 
base and the point D the vertex is a third of the prism which 
has the same base, the triangle ABC, and DBF as its 
opposite. 

Porism. From this it is manifest that any pyramid is a 
third part of the prism which has the same base with it and 
equal height 

o.. E. D. 

If we denote by C-ABD a pyramid with vertex C and base ABD, Euclid's 
argument is easily followed thus. 

The CD ABED being bisected by BD, 

(pyramid C-ABD) = (pyramid C-DEB) |xil. 5] 

3 (pyramid DEBC). 
And, the E3 EBCF being bisected by EC, 

(pyramid DEBC) = (pyramid D-ECF). 
Thus (pyramid C-ABD) = (pyramid D-EBC) = (pyramid D-ECF), and 
these three pyramids make up the whole prism, so that each is one- third of the 
prism. 

And, since (pyramid C-ABD) a (pyramid D-ABC), 

(pyramid D-ABC) = \ (prism ABC, DEE). 

Proposition 8. 

Similar pyramids which have triangular bases are in the 
triplicate ratio of their corresponding sides. 

Let there be similar and similarly situated pyramids of 





396 BOOK XII fxir. 8 

which the triangles ABC, DEF. are the bases and the points 
G, H the vertices ; 

I say that the pyramid ABCG has to the pyramid DEFH 
the ratio triplicate of that which BC has to EF. 






For let the parallelepipedal solids BGML, EHQP be 
completed. 

Now, since the pyramid ABCG is similar to the pyramid 
DEFH, 

therefore the angle ABC is equal to the angle DEF, 

the angle GBC to the angle HEF, 

and the angle AUG to the angle DEB ; 

and, as AB is to DE, so is BC to EF, and BG to EH. 

And since, as AB is to DE, so is BC to EF, 

and the sides are proportional about equal angles, 

therefore the parallelogram BM is similar to the parallelo- 
gram EQ. 

For the same reason 

BN is also similar to ER, and BK to EO ; 

therefore the three parallelograms MB, BK, BN are similar 
to the three EQ, EO, ER. 

But the three parallelograms MB, BK, BN are equal and 
similar to their three opposkes, 

and the three EQ, EO, ER are equal and similar to their 

three opposkes. [xi. 14] 

Therefore the solids BGML, EHQP are contained by 
similar planes equal in multitude. 

Therefore the solid BGML is similar to the solid EHQP. 

But similar parallelepipedal solids are in the triplicate ratio 
of their corresponding sides. [xi. 33] 



xii. 8] PROPOSITION 8 397 

Therefore the solid BGML has to the solid EHQP the 
ratio triplicate of that which the corresponding side BC has to 
the corresponding side EF. 

But, as the solid BGML is to the solid EHQP, so is the 
pyramid ABCG to the pyramid DEFH, 
inasmuch as the pyramid is a sixth part of the solid, because 
the prism which is half of the parallelepipedal solid [xi. 28J is 
also triple of the pyramid. [xii. 7] 

Therefore the pyramid ABCG also has to the pyramid 
DEFH the ratio triplicate of that which BC has to EF. 

Q. E. D. 

Porism. From this it is manifest that similar pyramids 
which have polygonal bases are also to one another in the 
triplicate ratio of their corresponding sides. 

For, if they are divided into the pyramids contained in 
them which have triangular bases, by virtue of the fact that 
the similar polygons forming their bases are also divided into 
similar triangles equal in multitude and corresponding to the 
wholes fvi. 20], 

then, as the one pyramid which has a triangular base in the 
one complete pyramid is to the one pyramid which has a 
triangular base in the other complete pyramid, so also will all 
the pyramids which have triangular bases contained in the 
one pyramid be to all the pyramids which have triangular 
bases contained in the other pyramid [v, u], that is, the 
pyramid itself which has a polygonal base to the pyramid 
which has a polygonal base. 

But the pyramid which has a triangular base is to the 
pyramid which has a triangular base in the triplicate ratio of 
the corresponding sides ; 

therefore also the pyramid which has a polygonal base has to 
the pyramid which has a similar base the ratio triplicate of 
that which the side has to the side. 

It is at once proved that, the pyramids being similar, the parallelepipeds 
constructed as shown in the figure are also similar. 

Consequently, as these latter are In the triplicate ratio of their corre- 
sponding sides [xi. 33], so are the pyramids which are their sixth parts 
respectively {being one third of the respective prisms on the same bases, i.e. 
of the halves of the respective parallelepipeds, xi. 28). 

As the Porism is not used where Euclid might have been expected to use 
it {see note on xn. 12, p, 416), there is some reason to doubt its genuineness. 
P only has it in the margin, though in the first hand. 



39« 



BOOK XII 



[xii.9 



Proposition 9. 

In equal pyramids which have triangular bases the bases 
are reciprocally proportional to the heights ; and those pyramids 
in which the bases are reciprocally proportional to the heights 
are equal. 

For let there be equal pyramids which have the triangular 
bases ABC, DEF and vertices the points G, H ; 
I say that in the pyramids ABCG, DEFH the bases are 
reciprocally proportional to the heights, that is, as the base 
ABC is to the base DBF, so is the height of the pyramid 
DEFH to the height of the pyramid ABCG. 




For let the parallelepipedal solids SGML, EHQP be 
completed. 

Now, since the pyramid ABCG is equal to the pyramid 
DEFH, 

and the solid BGML is six times the pyramid ABCG. 

and the solid EHQP six times the pyramid DEFH, 

therefore the solid BGML is equal to the solid EHQP, 

But in equal parallelepipedal solids the bases are recipro- 
cally proportional to the heights ; [xi. 34] 

therefore, as the base BM is to the base EQ, so is the height 
of the solid EHQP to the height of the solid BGML. 

But, as the base BM is to EQ, so is the triangle ABC to 
the triangle DEF. [1. 34] 

Therefore also, as the triangle ABC is to the triangle 
DEF, so is the height of the solid EHQP to the height of 
the solid BGML. [v. u] 



xii. 9] PROPOSITION 9 399 

But the height of the solid EHQP is the same with the 
height of the pyramid DEFH, 

and the height of the solid BGML is the same with the 
height of the pyramid ABCG t 

therefore, as the base ABC is to the base DEF, so is the 
height of the pyramid DEFH to the height of the pyramid 
ABCG. 

Therefore in the pyramids ABCG, DEFH the bases are 
reciprocally proportional to the heights. 

Next, in the pyramids ABCG, DEFH let the bases be 
reciprocally proportional to the heights ; 
that is, as the base ABC is to the base DEF, so let the height 
of the pyramid DEFH be to the height of the pyramid 
ABCG ; 

F say that the pyramid ABCG is equal to the pyramid 
DEFH 

For, with the same construction, 
since, as the base ABC is to the base DEF, so is the height 
of the pyramid DEFH to the height of the pyramid ABCG, 

while, as the base ABC is to the base DEF, so is the 
parallelogram BM to the parallelogram EQ, 

therefore also, as the parallelogram BM is to the parallelogram 
EQ, so is the height of the pyramid DEFH to the height of 
the pyramid ABCG. [v. ti] 

But the height of the pyramid DEFH is the same with 
the height of the parallelepiped EHQP, 

and the height of the pyramid ABCG is the same with the 
height of the parallelepiped BGML ; 

therefore, as the base BM is to the base EQ, so is the height 
of the parallelepiped EHQP to the height of the parallelepi- 
ped BGML. 

But those parallelepipedal solids in which the bases are 
reciprocally proportional to the heights are equal ; [xi. 34] 

therefore the parallelepipedal solid BGML is equal to the 
parallelepipedal solid EHQP. 

And the pyramid ABCG is a sixth part of BGML, and 
the pyramid DEFH a sixth part of the parallelepiped 
EHQP; 



400 BOOK XII [xn. 9, 10 

therefore the pyramid ABCG is equal to the pyramid DEFH, 
Therefore etc 

Q. E. D. 

The volumes of the pyramids are respectively one sixth part of the volumes 
of the parallelepipeds described, as in the figure, on double the bases and with 
the same heights as the pyramids. 

I. Thus the parallelepipeds are equal if the pyramids are equal. 

And, the parallelepipeds being equal, their bases are reciprocally propor- 
tional to their heights ; [xi. 34] 
hence the bases of the equal pyramids (which are the halves of the bases of 
the parallelepipeds) are proportional to their heights. 

II. If the bases of the pyramids are reciprocally proportional to their 
heights, so are the bases of the parallelepipeds to their heights (since the bases 
of the parallelepipeds are double of the bases of the pyramids respectively). 

Consequently the parallelepipeds are equal. fxi. 34] 

Therefore their sixth parts, the pyramids, are also equal. 



Proposition 10. 

Any cone is a third part of the cylinder which has the same 
base with it and equal height. 

For let a cone have the same base, namely the circle 
ABCD, with a cylinder and equal 
height ; 

I say that the cone is a third part 
of the cylinder, that is, that the 
cylinder is triple of the cone. 

For if the cylinder is not triple 
of the cone, the cylinder will be 
either greater than triple or less 
than triple of the cone. 

First let it be greater than 
triple, 

and let the square ABCD be 
inscribed in the circle ABCD ; [iv. 6] 

then the square ABCD is greater than the half of the circle 
ABCD. 

From the square ABCD let there be set up a prism of 
squal height with the cylinder. 

Then the prism so set up is greater than the half of the 
cylinder, 




xii. 10] PROPOSITIONS 9, 10 401 

inasmuch as, if we also circumscribe a square about the circle 
ABCD[iv. 7], the square inscribed in the circle A BCD is half 
of that circumscribed about it, 

and the solids set up from them are parallelepipedal prisms of 
equal height, 

while parallelepipedal solids which are of the same height are 
to one another as their bases ; [xi. 3*] 

therefore also the prism set up on the square ABCD is half 
of the prism set up from the square circumscribed about the 
circle ABCD ; [cf. xi. 28, or xii. 6 and 7, Por.] 

and the cylinder is less than the prism set up from the square 
circumscribed about the circle ABCD; 

therefore the prism set up from the square ABCD and of 
equal height with the cylinder is greater than the half of the 
cylinder. 

Let the circumferences AB, BC, CD, DA be bisected at 
the points E, E, G, H, 

and let A £, EB, BE, EC, CG, GD, DH, HA be joined ; 

then each of the triangles AEB, BEC, CGD, DMA is greater 
than the half of that segment of the circle ABCD which is 
about It, as we proved before. [xii. a] 

On each of the triangles AEB, BEC, CGD, DMA let 
prisms be set up of equal height with the cylinder ; 

then each of the prisms so set up is greater than the half part 
of that segment of the cylinder which is about it, 

inasmuch as, if we draw through the points E, E, G, H 
parallels to AB, BC, CD, DA, complete the parallelograms 
on AB, BC, CD, DA, and set up from them parallelepipedal 
solids of equal height with the cylinder, the prisms on the 
triangles AEB, BEC, CGD, DMA are halves of the several 
solids set up ; 

and the segments of the cylinder are less than the parallelepi- 
pedal solids set up ; 

hence also the prisms on the triangles AEB, BEC, CGD, 
DHA are greater than the half of the segments of the 
cylinder about them. 

Thus, bisecting the circumferences that are left, joining 



40» BOOK XII [xn. 10 

straight lines, setting up on each of the triangles prisms of 
equal height with the cylinder, 

and doing this continually, 

we shall leave some segments of the cylinder which will be 
less than the excess by which the cylinder exceeds the triple 
of the cone. [x, i] 

Let such segments be left, and let them be AE, £B, BF, 
FC, CG, GD, DH, HA ; 

therefore the remainder, the prism of which the polygon 
AEBFCGDH is the base and the height is the same as that 
of the cylinder, is greater than triple of the cone. 

But the prism of which the polygon AEBFCGDH is the 
base and the height the same as that of the cylinder is triple 
of the pyramid of which the polygon AEBFCGDH is the 
base and the vertex is the same as that of the cone ; [xn. 7, Por.] 

therefore also the pyramid of which the polygon AEBFCGDH 
is the base and the vertex is the same as that of the cone is 
greater than the cone which has the circle A BCD as base. 

But it is also less, for it is enclosed by it: 

which is impossible. 

Therefore the cylinder is not greater than triple of the cone. 

I say next that neither is the cylinder less than triple of 
the cone, 

For, if possible, let the cylinder be less than triple of the 
cone , 

therefore, inversely, the cone is greater than a third part of 

the cylinder. 

Let the square ABCD be inscribed in the circle ABCD ; 

therefore the square ABCD is greater than the half of the 
circle ABCD. 

Now let there be set up from the square ABCD a pyramid 
having the same vertex with the cone ; 

therefore the pyramid so set up is greater than the half part 
of the cone, 

seeing that, as we proved before, if we circumscribe a square 



xii. 10] PROPOSITION 10 403 

about the circle, the square A BCD will be half of the square 
circumscribed about the circle, 

and if we set up from the squares parallelepipedal solids of 
equal height with the cone, which are also called prisms, the 
solid set up from the square A BCD will be half of that set up 
from the square circumscribed about the circle ; 
for they are to one another as their bases. [xi. 32] 

Hence also the thirds of them are in that ratio ; 
therefore also the pyramid of which the square A BCD is the 
base is half of the pyramid set up from the square circum- 
scribed about the circle. 

And the pyramid set up from the square about the circle 
is greater than the cone, 

for it encloses it. 

Therefore the pyramid of which the square A BCD is the 
base and the vertex is the same with that of the cone is 
greater than the half of the cone. 

Let the circumferences AB, BC, CD, DA be bisected at 
the points E, F, G, H, 

and let AE, EB, BF, FC, CG, GD, DH, HA be joined ; 
therefore also each of the triangles AEB, BFC, CGD, DHA 
is greater than the half part of that segment of the circle 
ABCD which is about it 

Now, on each of the triangles AEB, BFC, CGD, DHA 
let pyramids be set up which have the same vertex as the 
cone ; 

therefore also each of the pyramids so set up is, in the same 
manner, greater than the half part of that segment of the cone 
which is about it. 

Thus, by bisecting the circumferences that are left, joining 
straight lines, setting up on each of the triangles a pyramid 
which has the same vertex as the cone, 

and doing this continually, 

we shell leave some segments of the cone which will be less 
than the excess by which the cone exceeds the third part of 
the cylindei. [x. 1] 

Let such be left, and let them be the segments on AE, 
EB, BF, FC, CG, GD, DH, HA ; 



4o 4 BOOK XII Txn. io 

therefore the remainder, the pyramid of which the polygon 
AEBFCGDH is the base and the vertex the same with that 
of the cone, is greater than a third part of the cylinder. 

But the pyramid of which the polygon AEBFCGDH is 
the base and the vertex the same with that of the cone is a 
third part of the prism of which the polygon AEBFCGDH 
is the base and the height is the same with that of the 
cylinder ; 

therefore the prism of which the polygon AEBFCGDH is 
the base and the height is the same with that of the cylinder 
is greater than the cylinder of which the circle A BCD is the 
base. 

But it is also less, for it is enclosed by it : 
which is impossible. 

Therefore the cylinder is not less than triple of the cone. 

But it was proved that neither is it greater than triple ; 
therefore the cylinder is triple of the cone ; 
hence the cone is a third part of the cylinder. 

Therefore etc 

Q. E. D. 



We observe the use in this proposition of the term " parallel pi pedal 
prism," which recalls Heron's " parallelogrammic " or " parallel-sided prism." 

The course of the proof is exactly the same as in xn. a, except that an 
arithmetical fraction takes the place of a ratio which, being incommensurable, 
could only be expressed as a ratio. Consequently we do not need proportions 
in this proposition, as we did in xn. z, and shall again in xn. 1 1 , etc. 

Euclid exhausts the cylinder and cone respectively by setting up prisms 
and pyramids of the same height on the successive regular polygons inscribed 
in the circle which is the common base, viz. the square, the regular polygon 
of 8 sides, that of 16 sides, etc. 

If A B be the side of one polygon, we obtain two sides of the next by 
bisecting the arc ACB and joining A C, CB. Draw the 
tangent DE at C and complete the parallelogram 
A BED. 

Now suppose a prism erected on the polygon of 
which AB is a side, and of the same height as that of 
the cylinder. 

To obtain the prism of the same height on the next 
polygon we add all the triangular prisms of the same 
height on the bases A CB and the rest. 

Now the prism on ACB is half the prism of the 
same height on the CD ABED as base. 

[cf. xi. a8) 




xii. 10] PROPOSITION 10 405 

And the prism on O ABED includes, and is greater than, the portion of 
the cylinder standing on the segment ACB of the circle. 

The same thing is true in regard to the other sides of the polygon of 
which A B is one side. 

Thus the process begins with a prism on the square inscribed in the circle, 
which is more than half the cylinder, the next prism (with eight lateral faces) 
takes away more than half the remainder, and so on ; 

hence [x. 1], if we proceed far enough, we shall ultimately arrive at a prism 
leaving over portions of the cylinder together less than any assigned volume. 

The construction of pyramids on the successive polygons exhausts the cone 
in exactly the same way. 

Now, if the cone is not equal to one-third of the cylinder, it must be either 
greater or less. 

I. Suppose, if possible, that, V, O being their volumes respectively, 

Construct successive inscribed polygons in the bases and prisms on them 
until we arrive at a prism P leaving over portions of the cylinder together less 
than (0- 3^), i.e. such that 

But P is triple of the pyramid on the same base and of the same height ; 
and this pyramid is included by, and is therefore less than, V; 
therefore P<.$V. 

But, by construction, P> 3 V \ 

which is impossible. 

Therefore O > 3 V. 

II. Suppose, if possible, that O <. 3 V. 
Therefore V> \0. 

Construct successive pyramids in the cone in the manner described until 
we arrive at a pyramid n leaving over portions of the cone together less than 
(V-\0\ i.e. such that 

Now 11 is one-third of the prism oh the same base and of the same height; 
and this prism is included by, and is therefore less than, the cylinder ; 
therefore n < \0. 

But, by construction, II > \Oi 

which is impossible. 

Therefore is neither greater nor less than 3 V, so that 

= $V. 

It will be observed that here, as in xii. 2, Euclid always ex/musts the solid 
by (as it were) building up to it from inside. Hence the solid to be exhausted 
must, with him, be supposed greater than the solid to which it is to be proved 
equal ; and this is the reason why, in the second part, the initial supposition 
is turned round. 

In this case too Euclid might have approximated to the cone and cylinder 
by circumscribing successive pyramids and prisms in the way shown, after 
Archimedes, in the note on xn. z. 



4 o6 



BOOK XII 



[xn. it 






Proposition ii. 



Cones and cylinders which are of the same height are to 
one another as their bases. 

Let there be cones and cylinders of the same height, 

let the circles A BCD, EFGH be their bases, XL, MN their 
axes and AC, EG the diameters of their bases ; 

I say that, as the circle A BCD is to the circle EFGH, so is 
the cone AL to the cone EN. 




For, if not, then, as the circle A BCD is to the circle 
EFGH, so will the cone AL be either to some solid less 
than the cone EN" or to a greater. 

First, let it be in that ratio to a less solid 0, and let the 
solid X be equal to that by which the solid O is less than the 
cone EN; 

therefore the cone EN is equal to the solids O, X. 

Let the square EFGH he inscribed in the circle EFGH; 

therefore the square is greater than the half of the circle- 
Let there be set up from the square EFGH a pyramid of 

equal height with the cone ; 

therefore the pyramid so set up is greater than the halfofthe 
cone, 

inasmuch as, if we circumscribe a square about the circle, and 
set up from it a pyramid of equal height with the cone, the 

inscribed pyramid is half of the circumscribed pyramid, 

for they are to one another as their bases, [xn. 6] 

while the cone is iess than the circumscribed pyramid. 



xii. nj PROPOSITION ii 407 

Let the circumferences EF, FG, GH, HE be bisected at 
the points P, Q, R, S, 
and let HP, PE, EQ, QF, FR, FG, GS, SH be joined. 

Therefore each of the triangles HPE, EQF, FRG, GSH 
is greater than the half of that segment of the circle which is 
about it. 

On each of the triangles HPE, EQF, FRG, GSH let 
there be set up a pyramid of equal height with the cone ; 
therefore, also, each of the pyramids so set up is greater than 
the half of that segment of the cone which is about it. 

Thus, bisecting the circumferences which are left, joining 
straight lines, setting up on each of the triangles pyramids of 
equal height with the cone, 
and doing this continually, 

we shall leave some segments of the cone which will be less 
than the solid X. [x. 1] 

Let such be left, and let them be the segments on HP, 
PE. EQ, QF, FR, RG, GS, SH; 

therefore the remainder, the pyramid of which the polygon 
HPFQFRGS is the base and the height the same with that 
of the cone, is greater than the solid O. 

Let there also be inscribed in the circle ABCD the 
polygon D TA UBVCW similar and similarly situated to the 
polygon HPEQFRGS, 

and on it let a pyramid be set up of equal height with the cone 
AL. 

Since then, as the square on AC is to the square on EG, so 
is the polygon DTA UBVCW to the polygon HPEQFRGS, 

[xii. 1] 

while, as the square on AC is to the square on EG, so is the 
circle ABCD to the circle EFGH, [xii. a] 

therefore also, as the circle ABCD is to the circle EFGH, so 
is the polygon D TA UB VCW to the polygon HPEQFRGS. 
But, as the circle ABCD is to the circle EFGH, so is the 
cone AL to the solid 0, 

and, as the polygon DTAUBVCW is to the polygon 
HPEQFRGS, so is the pyramid of which the polygon 
DTAUBVCW 'is the base and the point L the vertex to the 
pyramid of which the polygon HPEQFRGS is the base and 
the point A T the vertex. [w, 6] 



4 oS BOOK XII [xii. ii 

Therefore also, as the cone AL is to the solid O, so is the 
pyramid of which the polygon DTA UBVCW is the base and 
the point L the vertex to the pyramid of which the polygon 
HPEQFRGS is the base and the point N the vertex ; |v. 1 1] 
therefore, alternately, as the cone AL is to the pyramid in it, 
so is the solid O to the pyramid in the cone EN. [v. i6| 

But the cone AL is greater than the pyramid in it ; 
therefore the solid O is also greater than the pyramid in the 
cone EN. 

But it is also less: 
which is absurd. 

Therefore the cone AL is not to any solid less than the 
cone EN as the circle A BCD is to the circle EFGH. 

Similarly we can prove that neither is the cone EN to 
any solid less than the cone AL as the circle EFGH is to the 
circle ABCD. 

I say next that neither is the cone AL to any solid greater 
than the cone EN as the circle ABCD is to the circle 
EFGH. 

For, if possible, let it be in that ratio to a greater solid O \ 

therefore, inversely, as the circle EFGH is to the circle 
ABCD, so is the solid O to the cone AL. 

But, as the solid O is to the cone AL, so is the cone EN 
to some solid less than the cone AL ; 

therefore also, as the circle EFGH is to the circle ABCD, so 
is the cone EN to some solid less than the cone AL : 

which was proved impossible. 

Therefore the cone AL is not to any solid greater than 
the cone EN as the circle ABCD is to the circle EFGH. 

But it was proved that neither is it in this ratio to a less 
solrd ; 

therefore, as the circle ABCD is to the circle EFGH, so is 
the cone AL to the cone EN. 

But, as the cone is to the cone, so is the cylinder to the 
cylinder, 
for each is triple of each ; [xii. 10] 



xii. n] PROPOSITION ii 409 

Therefore also, as the circle ABCD is to the circle 
EFGH, so are the cylinders on them which are of equal 
height 

Therefore etc. 

Q. E. D. 

We need not again repeat the preliminary construction of successive 
pyramids and prisms exhausting the cones and cylinders. 

Let Z, Z" he the volumes of the two cones, ft ft their respective bases. 

If P-.P*Z:Z', 

then must $;$ = Z\Q, 

where O is either less or greater than Z'. 

I. Suppose, if possible, that is less than Z . 

Inscribe in Z' a pyramid (II'} leaving over portions of it together less than 
(Z' - O), i.e. such that 

Z' =* W > 0. 

Inscribe in Z a. pyramid II on a polygon inscribed in the circular base of 
Z similar to the polygon which is the base of IT. 
Now, if d, d' be the diameters of the bases, 

J9 : /T = ^ : rf'' [xii. 1] 

= (polygon in ft):{polygon in ft) [xn. 1] 

= 11:11'. [xii. 6] 

Therefore Z: = 11:11', 

and, alternately, Z: U = : II'. 

But "/, > n, since it includes it ; 

therefore 0> II'. 

But, by construction, <W : 

which is impossible. 

Therefore 0<fcZ 

II. Suppose, if possible, that 

yJ:ft = Z:0, 
where O is greater than Z'. 

Therefore : ft = O : Z\ 

where is some solid less than Z 

That is, ft : = Z' : 0', 

where O < Z 

This is proved impossible exactly in the same way as the assumption in 
Part I. was proved impossible. 

Therefore Z has not either to a less solid than Z' or to a greater solid than 
Z' the ratio of fl to ft j 

therefore 0:ft = Z:Z'. 

The same is true of the cylinders which are equal to $Z, $Z' respectively. 



4™ 









BOOK XII 



Proposition i 2. 



[x 11. ii 



Similar cones and cylinders are to one another in the 
triplicate ratio of the diameters in their bases. 

Let there be similar cones and cylinders, 
let the circles ABCD, EFGH be their bases, BD, FH the 

diameters of the bases, and KL t MN the axes of the cones 
and cylinders ; 

I say that the cone of which the circle ABCD is the base and 
the point L the vertex has to the cone of which the circle 
EFGH is the base and the point N the vertex the ratio 
triplicate of that which BD has to FH. 







H 




For, if the cone ABCDL has not to the cone EFGHN 
the ratio triplicate of that which BD has to FH, 

the cone ABCDL will have that triplicate ratio either to 
some solid less than the cone EFGHN or to a greater. 

First, let it have that triplicate ratio to a less solid O. 

Let the square EFGH be inscribed in the circle EFGH; 

[iv. 6] 
therefore the square EFGH is greater than the half of the 
circle EFGH. 

Now let there be set up on the square EFGH a pyramid 
having the same vertex with the cone ; 

therefore the pyramid so set up is greater than the half part 
of the cone. 



xti. it] PROPOSITION i3 411 

Let the circumferences EF, FG, GH, HE be bisected at 
the points P, Q, R, S, 
and let EP, PF, FQ, QG, GR, RH, HS, SE be joined. 

Therefore each of the triangles EPF, F JG, GRH, HSE 
is also greater than the half part of that segment of the circle 
EFGH which is about it. 

Now on each of the triangles EPF, FQG, GRH, HSE 
let a pyramid be set up having the same vertex with the cone; 
therefore each of the pyramids so set up is also greater than 
the half part of that segment of the cone which is about it. 

Thus, bisecting the circumferences so left, joining straight 
lines, setting up on each of the triangle- pyramids having the 
same vertex with the cone, 
and doing this continually, 

we shall leave some segments of the cone which will be less 
than the excess by which the cone EFGHN exceeds the 
solid O. [x. 1] 

Let such be left, and let them be the segmet: i on EP, 
PF, FQ, QG, GR, RH, HS, SE ; 

therefore the remainder, the pyramid of whirh the polygon 
EPFQGRHS is the base and the point N the vertex, is 
greater than the solid O. 

Let there be also inscribed in the circle ABCD the 
polygon ATBUCVDW similar and similarly situated to the 
polygon EPFQGRHS, 

and let there be set up on the polygon ATBUCVDW & 
pyramid having the same vertex with the cone ; 

of the triangles containing the pyramid of which the polygon 
ATBUCVDW is the base an 1 'ie point L the vertex let 
LB The one, 

and of the triangles containing the pyramid of which the 
polygon EPFQGRHS is the base and the point N the vertex 
let NFP be one ; 
and let KT, MP be joined. 

Now, since the cone ABCDL is similar to the cone 
EFGHN, 

therefore, as BD is to FH, so is the axis KL to th axis MN. 

[xi. Def. 24] 



4ia BOOK XII [xii. la 

But, as BD is to FN, so is BK to FM; 
therefore also, as BK is to FM, so is KL to MN. 

And, alternately, as BK is to AX, so is FM to ■#/!#. 

[v. ,6] 

And the sides are proportional about equal angles, namely 
the angles BKL, FMN ; 
therefore the triangle BKL is similar to the triangle FMN.. 

[vl 6] 

Again, since, as BK is to K T, so is FM to MP, 
and they are about equal angles, namely the angles BKT, 
FMP, 

inasmuch as, whatever part the angle BKT is of the four 
right angles at the centre K, the same part also is the angle 
FMP of the four right angles at the centre M ; 
since then the sides are proportional about equal angles, 
therefore the triangle BKT is similar to the triangle FMP. 

[vi. 6] 

Again, since it was proved that, as BK is to KL, so is FM 

to MN, 

while BK is equal to KT, and FM to PM, 
therefore, as TK is to KL, so is PM to MN ; 
and the sides are proportional about equal angles, namely 
the angles TKL, PMN, for they are right; 
therefore the triangle LKT is similar to the triangle NMP. 

[vi. 6] 

And since, owing to the similarity of the triangles LKB, 
NMF, 

as LB is to BK, so is NF to FM, 

and, owing to the similarity of the triangles BKT, FMP, 
as KB is to B T, so is MF to FP, 
therefore, ex aequali, as LB is to BT, so is NF to FP, [v. as] 

Again since, owing to the similarity of the triangles L TK, 
NPM, 

as L T is to TK, so is NP to PM, 

and, owing to the similarity of the triangles TKB, PMF, 
as KT is to TB, so is MP to ^ ; 
therefore, ex aequali, as LT is to 7!#, so is AT/ 3 to /^. [v. m] 



xii. i2] PROPOSITION la 413 

But it was also proved that, as TB is to BL, so is PF 
to FN. 

Therefore, ex aequali, as TL is to LB, so is PN to NF. 

[v. ix] 

Therefore in the triangles LTB, NPF the sides are 

proportional ; 

therefore the triangles LTB, NPF arc equiangular; [vi. 5] 

hence they are also similar, [vi. Def. 1] 

Therefore the pyramid of which the triangle BKT is the 
base and the point L the vertex is also similar to the pyramid 
of which the triangle FMP is the base and the point A'' the 
vertex, 

for they are contained by similar planes equal in multitude. 

[xi. Def. 9] 

But similar pyramids which have triangular bases are to 
one another in the triplicate ratio of their corresponding sides. 

[xii. 8] 

Therefore the pyramid BKTL has to the pyramid FMPN 
the ratio triplicate of that which BK has to FM. 

Similarly, by joining straight lines from A, IV, D, V, C, U 
to K, and from F, S, H, R, G, Q to M, and setting up on 
each of the triangles pyramids which have the same vertex 
with the cones, 

we can prove that each of the similarly arranged pyramids 
will also have to each similarly arranged pyramid the ratio 
triplicate of that which the corresponding side BK has to the 
corresponding side FM, that is, which BD has to FH. 

And, as one of the antecedents is to one of the conse- 
quents, so are all the antecedents to all the consequents ; 

[v. 12] 
therefore also, as the pyramid BKTL is to the pyramid 
FMPN, so is the whole pyramid of which the polygon 
ATBUCVDW'xs the base and the point L the vertex to the 
whole pyramid of which the polygon EPFQGRHS is the 
base and the point N the vertex ; 

hence also the pyramid of which A TB UCVD IV is the base 
and the point L the vertex has to the pyramid of which the 
polygon EPFQGRHS is the base and the point N the 
vertex the ratio triplicate of that which BD has to FH. 

But, by hypothesis, the cone of which the circle ABCD 



4U BOOK XII [xii. is 

is the base and the point L the vertex has also to the solid 
the ratio triplicate of that which BD has to FH; 

therefore, as the cone of which the circle ABCD is the base 
and the point L the vertex is to the solid O, so is the pyramid 
of which the polygon ATBUCVDW is the base and L the 
vertex to the pyramid of which the polygon EPFQGRHS is 
the base and the point N the vertex ; 

therefore, alternately, as the cone of which the circle ABCD 
is the base and L the vertex is to the pyramid contained in 
it of which the polygon A TB UC VD W is the base and L 
the vertex, so is the solid O to the pyramid of which the 
polygon EPFQGRHS is the base and N the vertex, [v. 16] 

But the said cone is greater than the pyramid in it ; 

for it encloses it. 

Therefore the solid O is also greater than the pyramid of 
which the polygon EPFQGRHS is the base and N the 
vertex. 

But it is also less : 

which is impossible. 

Therefore the cone of which the circle ABCD is the base 
and /, the vertex has not to any solid less than the cone of 
which the circle EFGH is the base and the point N the 
vertex the ratio triplicate of that which BD has to FH: 

Similarly we can prove that neither has the cone EFGHN 
to any solid less than the cone ABCDL the ratio triplicate 
of that which FH has to BD. 

I say next that neither has the cone ABCDL to any 
solid greater than the cone EFGHN the ratio triplicate of 
that which BD has to FH, 

For, if possible, let it have that ratio to a greater solid 0. 

Therefore, inversely, the solid has to the cone ABCDL 
the ratio triplicate of that which FH has to BD. 

But, as the solid O is to the cone ABCDL, so is the 
cone EFGHN to some solid less than the cone ABCDL. 

Therefore the cone EFGHN also has to some solid less 
than the cone ABCDL the ratio triplicate of that which FH 
has to BD : 
which was proved impossible. 



xii. u] PROPOSITION is 415 

Therefore the cone ABCDL has not to any solid greater 
than the cone EFGHN the ratio triplicate of that which BD 
has to FH. 

But it was proved that neither has it this ratio to a less 
solid than the cone EFGHN. 

Therefore the cone ABCDL has to the cone EFGHN 
the ratio triplicate of that which BD has to FH. 

But, as the cone is to the cone, so is the cylinder to the 
cylinder, 

for the cylinder which is on the same base as the cone and 
of equal height with it is triple of the cone; [xn. 10J 

therefore the cylinder also has to the cylinder the ratio 
triplicate of that which BD has to FH. 
Therefore etc. 

Q. E. D. 

The method of proof is precisely that of the previous proposition. The 
only addition is caused by the necessity of proving that, if similar equilateral 
polygons be inscribed in the bases of two similar cones, and pyramids be 
erected on them with the same vertices as those of the cones, the pyramids 
(are similar and) are to one another in the triplicate ratio of corresponding 
edges. 

Let KL, MJVba the axes of the cones, L, iVthe vertices, and let BT, FP 
be sides of similar polygons inscribed in the bases. Join BK, TK, BL, TL, 
PM, FM, PN, FN. 



I ^ / '[^% J 



Now BKL, FMN are right-angled triangles, and, since the cones are 
similar, 

BK : KL = FM : MN. [xi. Def. 14] 

Therefore (1) As BKL, FMN are similar. [vi. 6] 

Similarly (2) As TKL, PMNxk, similar. 

Next, in As BKT, FMP, the angles BKT, FMP are equal, since each is 
the same fraction of four right angles ; and the sides about the equal angles are 
proportional ; 
therefore (3) As BKT, FMP are similar. 





4i 6 BOOK XII [xii. i a 

Again, since from the similar A s BKL, FMN, and the similar A s BKT, 
FMP respectively, 

LB : BK = NF ': FM, 

BK:BT=MF:FP, 

ex aequali, LB : BT= NF: FP. 

Similarly LT : TB = NP ; PF. 

Inverting the latter ratio and compounding it with the preceding one, we 
have, ex aequali, 

LB:LT= NF: NP. 

Thus in As LTB, NPFthe sides are proportional in pairs; 
therefore {4) As LTB, NPFate similar. 

Thus the partial pyramids L-BKT, N-FMP are similar. 

In exactly the same way it is proved that all the other partial pyramids are 
similar. 

Now 
(pyramid L-BKT) ! {pyramid N-FMP) = ratio triplicate of ( BK: FM). 

The other partial pyramids are to one another in the same triplicate ratio. 

The sum of the antecedents is therefore to the sum of the consequents in 
the same triplicate ratio, 
i.e. (pyramid L-A TBU...): (pyramid N-EPFQ.. . ) 

= ratio triplicate of ratio (BK: FM) 
= ratio triplicate of ratio (BD : FH). 

[The fact that Euclid makes this transition from the partial pyramids to 
the whole pyramids in the body of this proposition seems to me to suggest 
grave doubts as to the genuineness of the Porism to xu. 8, which contains a 
similar but rather more general extension from the case of triangular pyramids 
to pyramids with polygonal bases. Were that Porism genuine, Euclid would 
have been more likely to refer to it than to repeat here the same arguments 
which it contains,] 

Now we are in a position to apply the method of exhaustion. 

If A", X' be the volumes of the cones, d, d' the diameters of their bases, and if 
(ratio triplicate of d : d') * X : X', 
then must (ratio triplicate of d : d') = X : O, 

where O is either less or greater than X'. 

I. Suppose that O is lets than X'. 

Construct in the way described a pyramid (IT) in X' leaving over portions 
of X' together less than (Jf - O), so that X ' > W > O, 
and construct in X a pyramid (II), with the same vertex as X has, on a 
polygon inscribed in its base similar to the base of IT. 

Then, by what has just been proved, 

II : n' = (ratio triplicate of d : d') 
= X:0, by hypothesis, 
and, alternately, U:X=W:0. 

But X includes, and is therefore greater than, II ; 
therefore O > n'. 

But, by construction, < U' : 

which is impossible. . 

Therefore cannot be less than X'. 



xii. i», 13] PROPOSITIONS 12, 13 417 

II. Suppose, if possible, that 

(ratio triplicate of d-.d') = X: O, 
where is greater than X' ; 

then (ratio triplicate of d : d' ) = Z : X ', 

or, inversely, (ratio triplicate of d' : d) = X' : Z, 

where Z is some solid less than X. 

This is proved impossible by the exact method of Part I. 

Hence O cannot be either greater or less than X', 
and X : X' = (ratio triplicate of ratio d : d'). 

Proposition 13. 

If a cylinder be cut by a plane which is parallel to its 
apposite planes, then, as the cylinder is to the cylinder, so will 
the axis be to the axis. 

For let the cylinder AD be cut by the plane GH which 
is parallel to the opposite planes AB, CD, 
and let the plane GH meet the axis at the point K\ 
I say that, as the cylinder BG is to the cylinder GD, so is 
the axis EK to the axis KF. 




For let the axis EF be produced in both directions to the 
points L, M, 

and let there be set out any number whatever of axes EN, NL 
equal to the axis EK, 

and any number whatever FO, OM equal to FK ; 
and let the cylinder PW on the axis LM be conceived of 
which the circles PQ, VW me the bases. 

Let planes be carried through the points N, O parallel to 
AB, CD and to the bases of the cylinder PW, 
and let them produce the circles RS, TU about the centres 
N, 0. 

Then, since the axes LN, NE, EK are equal to one 
another, 



4i8 BOOK XU [xit. 13 

therefore the cylinders QR, RB, BG are to one another as 
their bases. [xu. 1 1] 

But the bases are equal ; 
therefore the cylinders QR, RB, BG are also equal to one 

another. 

Since then the axes LN, NE, EK are equal to one 
another, 

and the cylinders QR, RB, BG are also equal to one another, 
and the multitude of the former is equal to the multitude of 
the latter, 

therefore, whatever multiple the axis KL is of the axis EK, 
the same multiple also will the cylinder QG be of the 
cylinder GB. 

For the same reason, whatever multiple the axis MK is 
of the axis KF, the same multiple also is the cylinder WG 
of the cylinder GD. 

And, if the axis KL is equal to the axis KM, the cylinder 
QG will also be equal to the cylinder GW, 

if the axis is greater than the axis, the cylinder will also be 

greater than the cylinder, 
and if less, less. 

Thus, there being four magnitudes, the axes EK, KF 
and the cylinders BG, GD, 

there have been taken equimultiples of the axis EK and of 
the cylinder BG, namely the axis Z.A'and the cylinder QG, 
and equimultiples of the axis KF and of the cylinder GD, 
namely the axis KM and the cylinder GW ; 
and it has been proved that, 

if the axis KL is in excess of the axis KM, the cylinder QG 
is also in excess of the cylinder GW, 

if equal, equal, 
and if less, less. 

Therefore, as the axis EK is to the axis KF, so is the 
cylinder BG to the cylinder GD. [v. Def. 5] 

Q. E. D. 

It is not necessary to reproduce the proof, as it follows exactly the method 
of vi. 1 and xi. as. 

The fact that cylinders described about axes of equal length and having 





xii. i3, m] PROPOSITIONS 13, 14 419 

equal bases are equal is inferred from xii. 11 to the effect that cylinders of 
equal height are to one another as their bases. 

That, of two cylinders with unequal axes but equal bases, the greater is 
that which has the longer axis is of course obvious either by application or by 
cutting off from the cylinder with the longer axis a cylinder with an axis of the 
same length as that of the other given cylinder. 

Proposition 14, 

Cones and cylinders which are on equal bases are to one 
anotker as their heights. 

For let EB, FD be cylinders on equal bases, the circles 
AB, CD ; 

I say that, as the cylinder EB is 
to the cylinder FD, so is the axis 
GH to the axis KL. 

For let the axis KL be pro- 
duced to the point N, 
let LN be made equal to the axis aITTOb riiTiM 

GH, ^-^ ^^ 

and let the cylinder CM be conceived about LN as axis. 

Since then the cylinders EB, CM are of the same height, 
they are to one another as their bases. [xii. 1 1] 

But the bases are equal to one another ; 
therefore the cylinders EB, CM 'are also equal. 

And, since the cylinder FM has been cut by the plane 
CD which is parallel to its opposite planes, 
therefore, as the cylinder CM is to the cylinder FD, so is the 
axis LN to the axis KL, [xii. 13] 

But the cylinder CM is equal to the cylinder EB, 
and the axis LN to the axis GH; 

therefore, as the cylinder EB is to the cylinder FD, so is the 
axis GH to the axis KL. 

But, as the cylinder EB is to the cylinder FD, so is the 
cone ABG to the cone CDK, [xii. 10] 

Therefore also, as the axis GH is to the axis KL, so is 
the cone ABG to the cone CDK and the cylinder EB to the 
cylinder FD. q. e. d. 

No separate proposition corresponding to this is necessary in the case of 
parallelepipeds, for xi. 25 really contains the property corresponding to that in 
this proposition as well as the property corresponding to that in XI 1. 13. 



420 



BOOK XII 



[XH. 15 



Proposition 15. 

In equal cones and cylinders the bases are reciprocally 
proportional to the heights ; and those cones and cylinders in 
•which the bases are reciprocally proportional to the lieights are 
equal. 

Let there be equal cones and cylinders of which the circles 
ABCD, EFGH are the bases ; 
let AC, EG be the diameters of the bases, 
and KL, MN the axes, which are also the heights of the 
cones or cylinders ; 
let the cylinders AO, EP be completed. 

I say that in the cylinders AO, EP the bases are re- 
ciprocally proportional to the heights, 

that is, as the base ABCD is to the base EFGH, so is the 
height MN to the height KL. 




For the height LK is either equal to the height MN or 
not equal. 

First, let it be equal. 

Now the cylinder AO is also equal to the cylinder EP. 

But cones and cylinders which are of the same height are 
to one another as their bases ; [xii. 1 1] 

therefore the base ABCD is also equal to the base EFGH. 

Hence also, reciprocally, as the base ABCD is to the base 
EFGH, so is the height MN to the height KL. 

Next, let the height LK not be equal to MN, 
but let MN be greater ; 

from the height MN let QN be cut off equal to KL, 
through the point Q let the cylinder EP be cut by the plane 
TUS parallel to the planes of the circles EFGH, RP, 



xu. is] PROPOSITION 15 4Ji 

and let the cylinder ES be conceived erected from the circle 
EFGH as base and with height NQ. 

Now, since the cylinder AO is equal to the cylinder EP, 
therefore, as the cylinder AO is to the cylinder ES, so is the 
cylinder EP to the cylinder ES. [v. 7] 

But, as the cylinder AO is to the cylinder ES, so is the 
base ABCD to the base EFGH, 
for the cylinders AO, ES are of the same height ; [xu. 11] 

and, as the cylinder EP is to the cylinder ES, so is the height 
MN to the height QN, 

for the cylinder EP has been cut by a plane which is parallel 
to its opposite planes. [xu. 13] 

Therefore also, as the base ABCD is to the base EFGH, 
so is the height MN to the height QN. [v. n] 

But the height QN is equal to the height KL \ 
therefore, as the base ABCD is to the base EFGH, so is the 
height MN to the height KL. 

Therefore in the cylinders AO, EP the bases are re- 
ciprocally proportional to the heights. 

Next, in the cylinders AO, EP\et the bases be reciprocally 
proportional to the heights, 

that is, as the base ABCD is to the base EFGH, so let the 
height MN be to the height KL \ 

I say that the cylinder AOis equal to the cylinder EP. 

For, with the same construction, 

since, as the base ABCD is to the base EFGH, so is the 
height MN to the height KL, 

while the height KL is equal to the height QN, 

therefore, as the base ABCD is to the base EFGH, so is the 

height MN to the height QN 

But, as the base ABCD is to the base EFGH, so is the 
cylinder AO to the cylinder ES, 

for they are of the same height ; [xu. n] 

and, as the height MN is to QN, so is the cylinder EP to the 
cylinder ES; [xu. 13] 

therefore, as the cylinder A O is to the cylinder ES, so is the 
cylinder EP to the cylinder ES. [v. n] 



BOOK XII 



[xn. is 



Therefore the cylinder AO is equal to the cylinder EP. 
And the same is true for the cones also. 

Q. E. D. 

I. If the heights of the two cylinders are equal, and their volumes are 
equal, the bases are equal, since the latter are proportional to the volumes, 

fxn. u] 
If the heights are not equal, cut off from the higher cylinder a cylinder of 
the same height as the lower. 

Then, if LK, QNYx the equal heights, 
we have, by xir. n, 

(base ASCD) : (base EFGH) = (cylinder AO) : 

= (cylinder EP) I 









= MN 
= MN 



ON 
KL. 



i (cylinder ES) 
l (cylinder ES), 
by hypothesis, 

[in. 13] 



II. In the converse part of the proposition, Euclid omits the case where 
the cylinders have equal heights. In this case of course the reciprocal ratios 
are both ratios of equality ; the bases are therefore equal, and consequently the 
cylinders. 

If the heights are not equal, we have, with the same construction as before, 

(base A BCD) • (base EFGH) = MN : KL. 
But [xn. nj 

(base ABCD) : (base EFGH) - (cylinder AO) : (cylinder ES), 
and MN:KL = MN\QN 

m (cylinder EP) : (cylinder ES). [xn. 13] 
Therefore 

(cylinder AO) ■ (cylinder ES) = (cylinder EP) : (cylinder ES), 
and consequently (cylinder AO) = (cylinder EP). 

Similarly for the cones, which are equal to one-third of the cylinders 
respectively. 

Legendre deduces these propositions about cones and cylinders 
others which he establishes by a method similar 
to that adopted by him for the theorem of xn. a 
(see note on that proposition). 

The first (for the cylinder) is as follows. 

The volume of a cylinder is equal to the 
product of its base by its height. 

Suppose CA to be the radius of the base of 
the given cylinder, h its height. 

For brevity let us denote by (surf. CA) the 
area of the circle of which CA is the radius. 

If (surf. CA)x h is not the measure of the 
given cylinder, it will be the measure of a 
cylinder greater or less than it. 

L First let it be the measure of a less 
cylinder, that, for example, of which the circle with radius CD is the base, and 
h is the height 



from two 




xii. is, 1 6] PROPOSITIONS 15, 1 6 423 

Circumscribe about the circle with radius CD a regular polygon GHI ... 
such that its sides do not anywhere meet the circle with radius CA. [See note 
on xii. a, p. 393 above, for Legendre's lemma relating to this construction.] 

Imagine a prism erected on the polygon as base and with height h. 

Then (volume of prism) = (polygon GHI...) x h. 

[Legendre has previously proved this proposition, first for a parallelepiped 
(by transforming it into a rectangular one), then for a triangular prism (half of 
a parallelepiped of the same height), and lastly for a prism with a polygonal 
base.] 

But (polygon GHI. ..)< (surf. CA). 

Therefore (volume of prism) < (surf. CA) x A 

< (cylinder on circle of rad. CD), 
by hypothesis. 

But the prism is greater than the latter cylinder, since it includes it ; 
which is impossible. 

II. In order not to multiply figures let us, in this second cose, suppose 
that CD is the radius of the base of the given cylinder, and that (surf. CD) * h 
is the measure of a cylinder greater than it, e.g. a cylinder on the circle with 
radius CA as base and of height h. 

Then, with the same construction, 

(volume of prism) = (polygon GHI...) x h. 

And (polygon GHI...\t> (surf. CD\ 

Therefore (volume of prism) > (surf. CD) x k 

> (cylinder on surf. CA), by hypothesis. 

But the volume of the prism is also less than that cylinder, being included 
by it: 
which is impossible. 

Therefore (volume of cylinder) = (its base) x (its height). 

It follows as a corollary that 

Cylinders of the same height are to one another as their bases f_xn. 13], and 
cylinders on the same base are to one another as their heights [xti. 14]. 

Also 

Similar cylinders are as the cubes of their heights, or as the cubes of the 
diameters of their oases [Eucl. XII. 13]. 

For the bases are as the squares on their diameters ; and, since the 
cylinders are similar, the diameters of the bases are as their heights. 

Therefore the bases are as the squares on the heights, and the bases 
multiplied by the heights, or the cylinders themselves, are as the cubes of the 
heights. 

I need not reproduce Legendre's proofs of the corresponding propositions 
for the cone. 

Proposition 16. 

Given two circles about the same centre, to inscribe in the 
greater circle an equilateral polygon with an even number of 
sides which does not touch the lesser circle. 




4 2 4 BOOK XII [xn. 16 

Let A BCD, EFGH be the two given circles about the 
same centre K; 

thus it is required to inscribe in the 
greater circle ABCD an equilateral 
polygon with an even number of 
sides which does not touch the circle 
EFGH. 

For let the straight line BKD 
be drawn through the centre K, 

and from the point G let GA be 
drawn at right angles to the straight 
line BD and carried through to C ; 

therefore AC touches the circle EFGH. [in. 16, Por.] 

Then, bisecting the circumference BAD, bisecting the 
half of it, and doing this continually, we shall leave a circum- 
ference less than AD. [x, 1] 

Let such be left, and let it be LD ; 

from L let LM be drawn perpendicular to BD and carried 
through to N, 

and let LD, DN be joined ; 

therefore LD is equal to DN. [in. 3, 1. 4] 

Now, since LN is parallel to AC, 

and AC touches the circle EFGH, 

therefore LN does not touch the circle EFGH; 

therefore LD, DN are far from touching the circle EFGH. 

If then we fit into the circle ABCD straight lines equal 
to the straight line LD and placed continuously, there will 
be inscribed in the circle ABCD an equilateral polygon with 
an even number of sides which does not touch the lesser 
circle EFGH. q. e. f. 

It must be carefully observed that the polygon inscribed in the outer circle 
in this proposition is such that not only do its own sides not touch the inner 
circle, but also the chords, as LN, joining angular points next hut out to each 
other do not touch the inner circle either. In other words, the polygon is the 
second in order, not the first, which satisfies the condition of the enunciation. 
This is important, because such a polygon is wanted in the next proposition ; 
hence in that proposition the exact construction here given must be followed. 



XII. I?] 



PROPOSITIONS iS, r 7 



4*5 



Proposition 17. 

Given two spheres about the same centre, to inscribe in the 
greater sphere a polyhedral solid which does not touch the 
lesser sphere at its surface. 

Let two spheres be conceived about the same centre A \ 
thus it is required to inscribe in the greater sphere a poly- 
hedral solid which does not touch the lesser sphere at its 
surface. 







Let the spheres be cut by any plane through the centre ; 
then the sections will be circles, 

inasmuch as the sphere was produced by the diameter 
remaining fixed and the semicircle being carried round it ; 

[xi. Def. 14] 
hence, in whatever position we conceive the semicircle to be, 
the plane carried through it will produce a circle on the 
circumference of the sphere. 

And it is manifest that this circle is the greatest possible, 



4*6 BOOK XII [xii. 17 

inasmuch as the diameter of the sphere, which is of course 
the diameter both of the semicircle and of the circle, is greater 
than all the straight lines drawn across in the circle or the 
sphere. 

Let then BCDE be the circle in the greater sphere, 
and FGH the circle in the lesser sphere ; 
let two diameters in them, BD, CE. je drawn at right angles 
to one another ; 

then, given the two circles BCDE, FGH about the same 
centre, let there be inscribed in the greater circle BCDE an 
equilateral polygon with an even number of sides which does 
not touch the lesser circle FGH, 

let BK, KL, LM, ME be its sides in the quadrant BE. 
let KA be joined and carried through to N, 
let AO be set up from the point A at right angles to the 
plane of the circle BCDE, and let it meet the surface of the 
sphere at O, 

and through AO and each of the straight lines BD, KN let 
planes be carried ; 

they will then make greatest circles on the surface of the 
sphere, for the reason stated. 

Let them make such, 
and in them let BOD, KON be the semicircles on BD, KN. 

Now, since OA is at right angles to the plane of the circle 
BCDE, 

therefore all the planes through OA are also at right angles 
to the plane of the circle BCDE ; [xi. 18] 

hence the semicircles BOD, KON are also at right angles to 
the plane of the circle BCDE. 

And, since the semicircles BED, BOD, KON are equal, 
for they are on the equal diameters BD, KN, 
therefore the quadrants BE, BO, KO are also equal to one 
another. 

Therefore there are as many straight lines in the quadrants 
BO, KO equal to the straight lines BK, KL, LM, ME as 
there are sides of the polygon in the quadrant BE. 

Let them be inscribed, and let them be BP, PQ, QR, RO 
and KS, ST, TU, UO, 
let SP, TQ, UR be joined, 



xii. 17] PROPOSITION 17 4»7 

and from P, S let perpendiculars be drawn to the plane of the 
circle BCDE ; [xi. 1 1] 

these will fall on BD, KN, the common sections of the planes, 
inasmuch as the planes of BOD, KONare also at right angles 
to the plane of the circle BCDE. [c£ xi. Def. 4] 

Let them so fall, and let them be PV, SW, 
and let WVbe joined. 

Now since, in the equal semicircles BOD, KON, equal 
straight lines BP, KS have been cut off, 
and the perpendiculars PV, SWhaxe been drawn, 
therefore PV is equal to SW, and BV to KW. [in. 27, 1. 26] 

But the whole BA is also equal to the whole KA ; 
therefore the remainder VA is also equal to the remainder WA ; 
therefore, as BV\% to VA, so is A^to WA ; 
therefore WV\s parallel to KB. [vi. 2] 

And, since each of the straight lines PV, SW is at right 
angles to the plane of the circle BCDE, 
therefore PV is parallel to SW. [xi. 6] 

But it was also proved equal to it ; 
therefore WV, SP are also equal and parallel. [1. 33] 

And, since WV is parallel to SP, 
while WV is parallel to KB, 
therefore SP is also parallel to KB. [xi. 9] 

And BP, Adjoin their extremities; 
therefore the quadrilateral KBPS is in one plane, 
inasmuch as, if two straight lines be parallel, and points be 
taken at random on each of them, the straight line joining the 
points is in the same plane with the parallels. [xi. 7] 

For the same reason 
each of the quadrilaterals SPQ T, TQR U is also in one plane. 
But the triangle URO is also in one plane. [xi. 2] 

If then we conceive straight lines joined from the points 
P, S, Q, T, R, U to A, there will be constructed a certain 
polyhedral solid figure between the circumferences BO, KO, 
consisting of pyramids of which the quadrilaterals KBPS, 
SPQT, TQR U and the triangle URO are the bases and the 
point A the vertex. 



4*8 BOOK XII Txii. 17 

And, if we make the same construction in the case of each 
of the sides KL, LM, ME as in the case of BK, and further 
in the case of the remaining three quadrants, 
there will be constructed a certain polyhedral figure in- 
scribed in the sphere and contained by pyramids, of which 
the satd quadrilaterals and the triangle URO, and the others 
corresponding to them, are the bases and the point A the 
vertex. 

1 say that the said polyhedron will not touch the lesser 
sphere at the surface on which the circle FGH is. 

Let AX be drawn from the point A perpendicular to the 
plane of the quadrilateral KB PS, and let it meet the plane at 
the point X ; [xi. 11] 

let XB, XK be joined. 

Then, since AX is at right angles to the plane of the 
quadrilateral KBPS, 

therefore it is also at right angles to all the straight lines 
which meet it and are in the plane of the quadrilateral. 

[xi. Def. 3] 

Therefore AX is at right angles to each of the straight 
lines BX, XK. 

And, since AB is equal to AK, 

the square on AB is also equal to the square on AK. 

And the squares on AX, XB are equal to the square 
on AB, 

for the angle at X is right ; [1. 47] 

and the squares on AX, XK are equal to the square on AK. 

[U] 

Therefore the squares on A X, XB are equal to the squares 
on AX, XK. 

Let the square on AX be subtracted from each ; 

therefore the remainder, the square on BX, is equal to the 
remainder, the square on XK; 

therefore BX is equal to XK. 

Similarly we can prove that the straight lines joined 
from X to P, S are equal to each of the straight lines BX, 
XK. 



xii. i7] PROPOSITION 17 419 

Therefore the circle described with centre X and distance 
one of the straight lines XB, XK will pass through P, S also, 
and KB PS will be a quadrilateral in a circle. 

Now, since KB is greater than WV, 
while WV\s equal to SP, 
therefore KB is greater than SP. 

But KB is equal to each of the straight lines KS, BP\ 
therefore each of the straight lines KS, BP is greater than SP, 

And, since KB PS is a quadrilateral in a circle, 
and KB, BP, KS are equal, and PS less, 
and BX is the radius of the circle, 

therefore the square on KB is greater than double of the 
square on BX. 

Let KZ be drawn from K perpendicular to B V. 

Then, since BD is less than double of DZ, 
and, as BD is to DZ, so is the rectangle DB, BZ to the 
rectangle DZ, ZB, 

if a square be described upon BZ and the parallelogram on 
ZD be completed, 

then the rectangle DB, BZ is also less than double of the 
rectangle DZ, ZB, 

And, if KD be joined, 
the rectangle DB, BZ is equal to the square on BK, 

and the rectangle DZ, ZB equal to the square on KZ ; 

[in. 31, vi. 8 and Por.J 
therefore the square on KB is less than double of the square 
on KZ. 

But the square on KB is greater than double of the square 
on BX ; 
therefore the square on KZ is greater than the square on BX. 

And, since BA is equal to KA, 
the square on BA is equal to the square on AK. 

And the squares on BX, XA are equal to the square on BA, 
and the squares on KZ, ZA equal to the square on KA ; 

['• 47] 
therefore the squares on BX, XA are equal to the squares on 
KZ. ZA, 



430 BOOK XII [xii. 17 

and of these the square on KZ is greater than the square 
on BX\ 

therefore the remainder, the square on ZA, is less than the 

square on XA. 

Therefore AX is greater than AZ ; 
therefore AX is much greater than AG, 

And AX is the perpendicular on one base of the poly- 
hedron, 

and AG on the surface of the lesser sphere ; 

hence the polyhedron will not touch the lesser sphere on its 
surface. 

Therefore, given two spheres about the same centre, a 
polyhedral solid has been inscribed in the greater sphere 
which does not touch the lesser sphere at its surface. 

Q. E. F. 

Porism, But if in another sphere also a polyhedral solid 
be inscribed similar to the solid in the sphere BCDE, 

the polyhedral solid in the sphere BCDE has to the poly- 
hedral solid in the other sphere the ratio triplicate of that 
which the diameter of the sphere BCDE has to the diameter 
of the other sphere. 

For, the solids being divided into their pyramids similar 
in multitude and arrangement, the pyramids will be similar. 

But similar pyramids are to one another in the triplicate 
ratio of their corresponding sides ; [xn. 8, Pot.] 

therefore the pyramid of which the quadrilateral KBPS is 
the base, and the point A the vertex, has to the similarly 
arranged pyramid in the other sphere the ratio triplicate of 
that which the corresponding side has to the corresponding 
side, that is, of that which the radius AB of the sphere about 
A as centre has to the radius of the other sphere. 

Similarly also each pyramid of those in the sphere about 
A as centre has to each similarly arranged pyramid of those 
in the other sphere the ratio triplicate of that which AB has 
to the radius of the other sphere. 

And, as one of the antecedents is to one of the conse- 
quents, so are all the antecedents to all the consequents ; 

[v. 12] 



xii. 17] 



PROPOSITION 17 



431 



hence the whole polyhedral solid in the sphere about A as 
centre has to the whole polyhedral solid in the other sphere 
the ratio triplicate of that which AB has to the radius of the 
other sphere, that is, of that which the diameter BD has to 
the diameter of the other sphere. 

Q. E. D. 

This proposition is of great length and therefore requires summarising in 
order to make it easier to grasp. Moreover there are some assumptions in it 
which require to be proved, and some omissions to be supplied. The figure 
also is one of some complexity, and, in addition, the text and the figure treat 
two points Z and V, which are really one and the same, as different. 

The first thing needed is to know that all sections of a sphere by planes 
through the centre are circles and equal to one another (great circles or 
" greatest circles " as Euclid calls them, more appropriately). Euclid uses his 
definition of a sphere as the figure described by a semicircle revolving about 
its diameter. This of course establishes that alt planes through the particular 
diameter make equal circular sections ; but it is also assumed that the same 
sphere is generated by any other semicircle of the same size and with its 
centre at the same point. 












8^=55^ 






sr su 








Sr jf 








ff\ ,./T 




. 




F 


_xkil "^^- 












n™ ^"^^sii 


A 






B t^y\ 


:1 \P ,-"'' 






— ~" D 















The construction and argument of the proposition may be shortly given 
as follows. 

A plane through the centre of two concentric spheres cuts them in great 
circles of which BE, GEzie quadrants. 

A regular polygon with an even number of sides is inscribed {exactly as in 
Prop. 16) to the outer circle such that its sides do not touch the inner circle. 
BK, KL, LM, AfE are the sides in the quadrant BE. 



43* BOOK xn [m 17 

AO is drawn at right angles to the plane ABE, and through AO are 
drawn planes passing through B, K, L, M, E, etc., cutting the sphere in great 
circles. 

OB, OK are quadrants of two of these great circles. 

As these quadrants are equal to the quadrant BE, they wilt be divisible 
into arcs equal in number and magnitude to the arcs UK, KL, LM, ME. 

Dividing the other quadrants of these circles, and also all the quadrants of 
the other circles through CM, in this way we shall have in all the circles a 
polygon equal to that in the circle of which BE is a quadrant 

BP, PQ, QR, RO and KS, ST, TU, UO are the sides of these polygons 
in the quadrants BO, KO. 

Joining PS, QT, RU, and making the same construction all round the 
circles through AO, we have a certain polyhedron inscribed in the outer 
sphere. 

Draw PV perpendicular to AB and therefore (since the planes OAB, 
BAE are at right angles) perpendicular to the plane BAE; [xr. Def. 4 J 

draw SW perpendicular to AK and therefore (for a like reason) perpendicular 
to the plane BAE. 

Draw KZ perpendicular to BA. (Since BK =BP, and DB.BV=BP t , 
DB.BZ=BJP, it follows that BV=BZ, and Z, V coincide.) 

Now, since l. s PA V, SA W, being angles subtended at the centre by 
equal arcs of equal circles, are equal, 
and since L s PVA, SWA are right, 

while AS=AP t 

A s PA F, SAW are equal in all respects, [1. 26] 

and A V= A W. 

Consequently AB: AV = AK-.AW; 

and VW % BKzie parallel. 

But PV, SW are parallel (being both perpendicular to one plane) and 
equal (by the equal As PA V, SA W), 
therefore VW % PS are equal and parallel. 

Therefore BK (being parallel to VW) is parallel to PS. 

Consequently (1) BPSK is a quadrilateral in one plane. 

Similarly the other quadrilaterals PQTS, QRl/T a.se in one plant ; and 
the triangle OR U is in one plane. 

In order now to prove that the plane BPSK does not anywhere touch the 
inner sphere we have to prove that the shortest distance from A to the plane 
is greater than AZ, which by the construction in XU. 16 is greater than AG, 

Draw AX perpendicular to the plane BPSK. 

Then AX* + XB* = A X* + XK' = AX'+XS , = AX t + XP* = AB 1 , 
whence XB = XK= XS = XP, 

or (2) the quadrilateral BPSK is inscri bable in a circle with X as centre and 
radius XB. 

Now BK> VW 

>PS; 
therefore in the quadrilateral BPSK three sides BK, BP, KS are equal, but 
PS is less. 

Consequently the angles about X are three equal angles and one smaller 
angle ; 



xii. 1 7] PROPOSITION ij 433 

therefore any one of the equal angles is greater than a right angle, i.e. u BXK 
is obtuse. 

Therefore (3) BK*>iBX*. [11. u] 

Next, consider the semicircle BKD with KZ drawn perpendicular to BD. 

We have BD < 2 DZ, 

so that D B . BZ < 2DZ ,-ZB, 

or BK - < iXZ 1 ; 

therefore, a fortiori, [by (3) above] 

(4) BX* < KZK 

Now AJO = AB t ; 

therefore AZ* + ZK* = AX % + XB>. 

And BX*<KZ*i 

therefore AX^AZ 1 , 

or (s) AX^AZ. 

But, by the construction in xii. 16, AZ>AG; therefore, a fortiori, 
AX>AG. 

And, since the perpendicular AX is the shortest distance from A to the 
plane BPSK, 
(6) the plane BPSK does not anywhere meet the inner sphere. 

Euclid omits to prove that, a fortiori, the other quadrilaterals PQTS, 
QRUT, and the triangle ROU, do not anywhere meet the inner sphere. 

For this purpose it is only necessary to show that the radii of the circles 
circumscribing BPSK, PQTS, QR UT and ROU we in descending order of 
magnitude. 



We have therefore to prove that, if A BCD, A' BCD' are two quadrilaterals 
inscribable in circles, and 

AD = BC=A'jy = B'C, 
while AB is not greater than AD, A'B = CD, and AB> CD> CV, 
then the radius OA of the circle circumscribing the first quadrilateral is greater 
than the radius OA' of the circle circumscribing the second, 

Clavius, and Simson after him, prove this by reductio ad absurdum. 

(1) HOA^&A', 
it follows that M 40D, BOC, A' OH, B'ffC are all equal. 
Also t-AOB^i.A'OB', 

LCOD^ucan', 

whence the four angles about O are together greater than the four angles 
about 0', i.e. greater than four right angles ; 
which is impossible. 



434 BOOK XII [xit. 17, 18 

(z) If a A' > OA, 
cut off from 0A\ US, OC\ O'D lengths equal to OA, and draw the inner 
quadrilateral as shown in the figure {XYZW). 

Then AB>A , ff>XY, 

CD>C'D>ZW, 
AD = A'JDr* WX, 
BC=SC> YZ 

Consequently the same absurdity as in (1) follows a fortiori. 

Therefore, since OA is neither equal to nor less than OA', 
OA > OA'. 

The fact is also sufficiently clear if we draw MO, NO bisecting DA, DC 
perpendicularly and therefore meeting in O, the centre of the circumscribed 
circle, and then suppose the side DA with the perpendicular MO to turn 
inwards about D as centre. Then the intersection of MO and NO, as P, will 
gradually move towards N. 

Simson gives his proof as " Lemma n." immediately before xit. 1 7. 
He adds to the Porism some words explaining how we may construct a 
similar polyhedron in another sphere and how we may prove that the 
polyhedra are similar. 

The Porism is of course of the essence of the matter because it is the 
porism which as much as the construction is wanted in the next proposition. 
It would therefore not have been amiss to include the Porism in the enuncia- 
tion of Xii. 1 7 so as to call attention to it 



Proposition 18. 

Spheres are to one another in the triplicate ratio of their 
respective diameters. 

Let the spheres ABC, DEFbe conceived, 
and let BC, EF be their diameters ; 

I say that the sphere ABC has to the sphere DEF the ratio 
triplicate of that which BC has to EF, 

Foi, if the sphere ABC has not to the sphere DEF the 
ratio triplicate of that which BC has to EF, 
then the sphere ABC will have either to some less sphere 
than the sphere DEF, or to a greater, the ratio triplicate of 
that which BC has to EF. 

First, let it have that ratio to a less sphere GHK, 
let DEF be conceived about the same centre with GHK, 
let there be inscribed in the greater sphere DEF a poly- 
hedral solid which does not touch the lesser sphere GHK at 
its surface, [xu. 17] 



XII. I 



8] 



PROPOSITIONS 17, 18 



43S 



and let there also be inscribed in the spnere ABC a poly- 
hedral solid similar to the polyhedral solid in the sphere DEF ; 

therefore the polyhedral solid in ABC has to the polyhedral 
solid in DEF the ratio triplicate of that which Z?Chas to EF. 

[xii, 17, Por.] 




But the sphere ABC also has to the sphere GHK the 
ratio triplicate of that which BC has to EF; 

therefore, as the sphere ABC is to the sphere GHK, so is 
the polyhedral solid in the sphere ABC to the polyhedral 
solid in the sphere DEF; 

and, alternately, as the sphere ABC is to the polyhedron in 
it, so is the sphere GHK to the polyhedral solid in the 
sphere DEF. [v. 16] 

But the sphere ABC is greater than the polyhedron in it ; 

therefore the sphere GHK is also greater than the polyhedron 
in the sphere DEF. 

But it is also less, 

for it is enclosed by it. 

Therefore the sphere ABC has not to a less sphere than 
the sphere DEF the ratio triplicate of that which the diameter 
BC has to EF. 



436 BOOK XII [xn. iS 

Similarly we can prove that neither has the sphere DEF 
to a less sphere than the sphere ABC the ratio triplicate of 
that which EF has to BC. 

I say next that neither has the sphere ABC to any greater 
sphere than the sphere DEF the ratio triplicate of that which 
BC has to EF. 

For, if possible, let it have that ratio to a greater, LMN; 

therefore, inversely, the sphere LMN has to the sphere ABC 
the ratio triplicate of that which the diameter EF has to the 
diameter BC. 

But, inasmuch as LMN is greater than DEF, 

therefore, as the sphere LMN is to the sphere ABC, so is the 
sphere DEF to some less sphere than the sphere ABC, as 
was before proved. [xn. a, Lemma] 

Therefore the sphere DEF also has to some less sphere 
than the sphere ABC the ratio triplicate of that which EF 
has to BC: 

which was proved impossible. 

Therefore the sphere ABC has not to any sphere greater 
than the sphere DEF the ratio triplicate of that which BC 
has to EF. 

But it was proved that neither has it that ratio to a less 
sphere. 

Therefore the sphere ABC has to the sphere DEF the 
ratio triplicate of that which BC has to EF. 

Q. E. D. 

It is the method of this proposition which Legendre adopted for his proof 
of xn. 2 (see note on that proposition). 

The argument can be put very shortly. We will suppose S, S' to be the 
volumes of the spheres, and d, d' to be their diameters ; and we will for brevity 
express the triplicate ratio of d to d' by d* : d''. 

ir rf*:^* s.s 1 , 

then J*:J*mS:T, 

where T is the volume of some sphere either greater or less than S'. 

I. Suppose, if possible, that 7"< S. 
Let T be supposed concentric with S'. 

As in xn. 17, inscribe a polyhedron in 5' such that its faces do not any- 
where touch T; 

and inscribe in S a polyhedron similar to that in S'. 



xii. i8] PROPOSITION 18 437 

Then S:T=d 1 -. rf' s 

■ (polyhedron in S) : (polyhedron in S') j 
or, alternately, 

S : (polyhedron in S) = T: {polyhedron in S'). 
And S > (polyhedron in S) ; 

therefore T> (polyhedron in 5'). 

But, by construction, T< (polyhedron in S'} ; 
which is impossible. 

Therefore T-$. S'. 

II. Suppose, if possible, that T > S'. 

Now d*;d' l = S: T 

= X\S\ 
where X is the volume of some sphere less than S, [xn. 2, Lemma] 

or, inversely, d' % : d* = S" : X, 

where X < S. 

This is proved impossible exactly as in Part 1. 

Therefore T $■ S. 

Hence T, not being greater or less than S', is equal to it, and 
d* : d* = S : S". 


















BOOK XIII. 

HISTORICAL NOTE. 

I have already given, in the note to iv. 10, the evidence upon which the 
construction of the live regular solids is attributed to the Pythagoreans. Some 
of them, the cube, the tetrahedron (which is nothing but a pyramid), and the 
octahedron (which is only a double pyramid with a square base), cannot but 
have been known to the Egyptians. And it appears that dodecahedra have 
been found, of bronze or other material, which may belong to periods earlier 
than Pythagoras' time by some centuries (for references see Cantor's Geschichte 
der Mathematik I,, pp, 175 — 6). 

It is true that the author of the scholium No. 1 to Eucl. xm. says that the 
Book is about "the five so-called Platonic figures, which however do not 
belong to Plato, three of the aforesaid five figures being due to the Pythagoreans, 
namely the cube, the pyramid and the dodecahedron, while the octahedron 
and the icosahed:on are due to Theaetetus." This statement (taken probably 
from Geminus) may perhaps rest on the fact that Theaetetus was the first to 
write at any length about the two last-mentioned solids. We are told indeed 
by Suidas (s. v. Sfamrro*} that Theaetetus " first wrote on the ' five solids ' as 
they are called." This no doubt means that Theaetetus was the first to write 
a complete and systematic treatise on all the regular solids ; it .does not 
exclude the possibility that Hippasus or others had already written on the 
dodecahedron. The fact that Theaetetus wrote upon the regular solids agrees 
very well with the evidence which we possess of his contributions to the 
theory of irrationals, the connexion between which and the investigation of 
the regular solids is seen in Euclid's Book xm. 

Theaetetus flourished about 380 B.C., and his work on the regular solids 
was soon followed by another, that of Aristaeus, an elder contemporary of 
Euclid, who also wrote an important book on Solid Loci, i.e. on conies treated 
as loci. This Aristaeus (known as "the elder") wrote in the period about 
310 B.C. We hear of his Comparison of the five regular solids from Hypsicles 
(2nd cent B.C.), the writer of the short book commonly included in the editions 
of the Elements as Book Xiv. Hypsicles gives in this Book some six proposi- 
tions supplementing Eucl. xm. ; and he introduces the second of the 
propositions (Heiberg's Euclid, Vol. v, p. 6) as follows : 

" The same circle circumscribes both the pentagon of the dodecahedron and the 
triangle of the icosahedron when both are inscribed in the same sphere. This is 
proved by Aristaeus in the book entitled Comparison #f the five figures" 



HISTORICAL NOTE 439 

Hypsicles proceeds (pp. 7 sqq.) to give a proof of this theorem. Allman 
pointed out {Greek Geometry from Thaies to Euclid, 1889, pp. 201 — 2) that this 
proof depends on eight theorems, six of which appear in Eudid's Book xm. 
(in Propositions 8, ro, 12, 15, 16 with Por., 17) ; two other propositions not 
mentioned by Allman are also used, namely xm. 4 and 9. This seems, as 
Allman says, to confirm the inference of Bretschneider (p. 171) that, as 
Aristaeus' work was the newest and latest in which, before Euclid's time, this 
subject was treated, we have in Eucl. xm. at least a partial recapitulation of 
the contents of the treatise of Aristaeus. 

After Euclid, Apollonius wrote on the comparison of the dodecahedron 
and the icosahedron inscribed in one and the same sphere. This we also 
learn from Hypsicles, who says in the next words following those about 
Aristaeus above quoted : "But it is proved by Apollonius in the second 
edition of his Comparison of the dodecahedron with the icosahedron that, as the 
surface of the dodecahedron is to the surface of the icosahedron [inscribed 
in the same sphere], so is the dodecahedron itself [i.e. its volume] to the 
icosahedron, because the perpendicular is the same from the centre of the 
sphere to the pentagon of the dodecahedron and to the triangle of the 
icosahedron." 









































































































BOOK XIII. PROPOSITIONS. 
Proposition i. 



If a straight line be cut in extreme and mean ratio, the 
square on the greater segment added to the half of the whole 
is five times the square on the half 

For let the straight line AB be cut in extreme and mean 
ratio at the point C, 
and let AC be the greater segment ; 
let the straight line AD be pro- 
duced in a straight line with CA, 
and let AD be made half of AB; 
F say that the square on CD is 
five times the square on AD. 

For let the squares AE, DF 
be described on AB, DC, 
and let the figure in DF be drawn ; 
let FC be carried through to G. 

Now, since AB has been cut in 
extreme and mean ratio at C, 
therefore the rectangle AB, BC is 
equal to the square on AC. 

[vi. Def. 3, vi. 17] 

And CM is the rectangle AB, BC, and FH the square 
on AC; 

therefore CE is equal to FH. 

And, since BA is double of AD, 
while BA is equal to KA, and AD to AH, 
therefore KA is also double of AH. 

But, as KA is to AH, so is CK to CH; 
therefore CK is double of CH. 

But LH, HC are also double of CH. 

Therefore KC is equal to LH, HC. 




[vi- 1] 



XIII. l] 



PROPOSITION i 



But CE was also proved equal to HF ; 
therefore the whole square AE is equal to the gnomon MNO. 

And, since BA is double of AD, 
the square on BA is quadruple of the square on AD, 
that is, AE is quadruple of DH. 

But AE is equal to the gnomon MNO ; 
therefore the gnomon MNO is also quadruple of AP; 
therefore the whole DF is five times AP. 

And DF is the square on DC, and AP the square on DA ; 
therefore the square on CD is five times the square on DA. 

Therefore etc, 

Q. E. D. 

The first five propositions are in the nature of lemmas, which are required 
for later propositions but are not in themselves of much importance. 

It will be observed that, while the method of the propositions is that of 
Book ii., being strictly geometrical and not algebraical, none of the results of 
that Book are made use of (except indeed in the Lemma to xin. a, which is 
probably not genuine). It would therefore appear as though these propositions 
were taken from an earlier treatise without being revised or rewritten in the 
light of Book 11. It will be remembered that, according to Froclus (p. 67, 6), 
Eudoxus " greatly added to the number of the theorems which originated with 
Plato regarding the section " {i.e. presumably the "go/den section ") ; and it is 
therefore probable that the five theorems are due to Eudoxus. 

That, if AB is divided at C in extreme and mean ratio, the rectangle 
AB, BC is equal to the square on AC is inferred from vi. 17. 

AD is made equal to half AB, and we have to prove that 
(sq. on CD) = 5 (sq. on AD). 

The figure shows at once that 

cdch^cdhl, s <- 

so that CJCH+njHL = i(CjCH) 
= £JAG. 
Also sq. HF= (sq. on AC) 

= rect AB, BC * ■ 

= CE. 
By addition, 

(gnomon MNO) - sq. on AB R 

= 4 (sq. on AD) ; 
whence, adding the sq. on AD to each, we have 
(sq. on CD) = 5 (sq. on AD). 
The result here, and in the next propositions, 
is really seen more readily by means of the figure 
of it. 11. 

In this figure SR = AC + ^AB, by construction; 

and we have therefore to prove that 

(sq. on SR) = s (sq. on AH). 




44* BOOK XIII [xm. i 

This is obvious, for 

(sq. on SR) = (sq. on RB) 

t AB AD 

- sum of sqs. on AB, AR 
= 5 (sq. on AR). 

The mss. contain a. curious addition to xin, i — 5 in the shape of analyses 
and syntheses for each proposition prefaced by the heading : 

" What is analysis and what is synthesis. 

" Analysis is the assumption of that which is sought as if it were admitted 
-; and the arrival > by means of its consequences at something admitted to 
be true. 

" Synthesis is an assumption of that which is admitted < and the arrival > 
by means of its consequences at something admitted to be true." 

There must apparently be some corruption in the text ; it does not, in the 
case of synthesis, give what is wanted. B and V have, instead of " something 
admitted to be true," the words " the end or attainment of what is sought." 

The whole of this addition is evidently interpolated. To begin with, the 
analyses and syntheses of the five propositions are placed all together in four 
mss. ; in P, q they come after an alternative proof of xui. 5 (which alternative 
proof P gives after xiit. 6, while q gives it instead of xm, 6), in B {which has 
not the alternative proof of xni. 5) after xm. 6, and in b (in which xm. 6 is 
wanting, and the alternative proof of xm. 5 is in the margin, in the first hand) 
after xjii. 5, while V has the analyses of 1 — 3 in the text after xm. 6 and 
those of 4 — 5 in the same place in the margin, by the second hand-. Further, 
the addition is altogether alien from the plan and manner of the Elements. 
The interpolation took place before The oil's time, and the probability is that 
it was originally in the margin, whence it crept into the text of P after xm. 5. 
Heiberg (after Bretschn eider) suggested in his edition (VoL v. p. Ixxxiv.) that 
it might be a relic of analytical investigations by Theaetetus or Eudoxus, and 
he cited the remark of Pappus (v. p. 410) at the beginning of his 
"comparisons of the five [regular solid] figures which have an equal surface," 
to the effect that he will not use " the so-called analytical investigation by 
means of which some of the ancients effected their demonstrations." More 
recently {Ptxralipomena zu Rukiid in Hermes xxxvm., 1903) Heiberg con- 
jectures that the author is Heron, on the ground that the sort of analysis and 
synthesis recalls Heron's remarks on analysis and synthesis in his commentary 
on the beginning of Book it. (quoted by an-NairizI, ed. Curtze, p. 89) and his 
quasi-algebraical alternative proofs of propositions in that Book. 

To show the character of the interpolated matter I need only give the 
and synthesis of one proposition. In the case of xni. 1 it is in 
substance as follows. The figure is a mere 
straight line. DA C 1 

Let A B be divided in extreme and mean 1 1 ' ' 

ratio at C, AC being the greater segment ; 
and let AD=\AB. 

I say that (sq. on CD) = 5 (sq. on AD). 

(Analysis.) 

" For, since (sq. on CD) = 5 (sq. on AD)" 

and (sq. on CD) = (sq. on CA) + (sq. on AD) + 2 (rect. CA, AD), 
therefore (sq. on CA) + a (rect. CA, AD) = 4 (sq. on AD). 

But rect. BA . A C = z (rect. CA . AD), 

and (sq. on CA) = (rect. AB, £C), 



Xlll. I, a] 



PROPOSITIONS 



443 



Therefore 

(rect BA, AC) + (rect. AB, BC) = 4 (sq. on AD), 
or (sq. on AB) = 4 (sq. on AD) : 

and this is true, since AD = \AB. 

(Synthesis.) 

Since (sq. on AB) = 4(sq. on AD), 

and (sq. on AB) = (rect. BA, ,4(7) + (rect. AB, BC), 

therefore 4(sq. on AD) = 2 (rect. DA, AC) + sq. on AC. 

Adding to each the square on AD, we have 

(sq. on CD) = 5 (sq. on AD). 

Proposition 2. 

If the square on a straight line be five times the square on 
a segment of it, then, when the double of the said segment is cut 
in extreme and mean ratio, the greater segment is the remaining 
part of the original straight line. 

For let the square on the straight line AB be five times 
the square on the segment AC 
of it, 

and let CD be double of AC ; 
I say that, when CD is cut in extreme 
and mean ratio, the greater segment 
is CB. 

Let the squares AF, CG be de- 
scribed on AB, CD respectively, 
let the figure in AFbe drawn, 
and let BE be drawn through. 

Now, since the square on BA is 
five times the square on AC, 
AF is five times AH. 

Therefore the gnomon MNO is 
quadruple of AH. 

And, since DC is double of CA, 
therefore the square on DC is quadruple of the square on CA, 
that is, CG is quadruple of AH. 

But the gnomon MNO was also proved quadruple of AH; 
therefore the gnomon MNO is equal to CG, 

And, since DC is double of CA, 
while DC is equal to CK, and AC to CH, 
therefore KB is also double of BH. [vi. 1]. 



L 




F 




u/ 




>** 






M- 


/\ 






H , ; 




f 









B D 


A 







444 BOOK XIII [xnr. i 

But LH, HB are also double of HB ; 
therefore KB is equal to LH, HB. 

But the whole gnomon MNO was also proved equal to 
the whole CG ; 
therefore the remainder HF is equal to BG. 

And BG is the rectangle CD, DB, 
for CD is equal to DG ; 
and HF is the square on CB \ 
therefore the rectangle CD, DB is equal to the square on CB. 

Therefore, as DC is to CB, so is CB to BD. 

But DC is greater than CB ; 
therefore CB is also greater than BD. 

Therefore, when the straight line CD is cut in extreme and 
mean ratio, CB is the greater segment. 

Therefore etc. 

Q. E. D. 

■ 
Lemma. 

That the double of A C is greater than BC is to be proved 
thus. 

If not, let BC be, if possible, double of CA. 

Therefore the square on BC is quadruple of the square 
on CA ; 

therefore the squares on BC, CA are five times the square 
on CA. 

But, by hypothesis, the square on BA is also five times 
the square on CA ; 

therefore the square on BA is equal to the squares on BC, CA : 
which is impossible. [n. 4] 

Therefore CB is not double of AC. 

Similarly we can prove that neither is a straight line less 
than CB double of CA ; 
for the absurdity is much greater. 

Therefore the double of A C is greater than CB. 

Q. E. D. 

This proposition is the converse of Prop. 1. We have to prove that, if 
AB be so divided at C that 

(sq. on AS) = 5 (sq. on AC), 
and if CD = a AC, 
then (rect. CD, DB) = (sq. on CB). 



h 3] 



PROPOSITIONS 2, 3 



US 



Subtract from each side the sq. on AC; 
then (gnomon MNO) = 4(sq. on AC) 

= (sq. on CD). 
Now, as in the last proposition, 

CJCE = 2(E) BH) 

= a BH '+ O HL. 

Subtracting these equals from the equals, the square on CD and the 
gnomon MNO respectively, we have 

a BG= (square HF), 

i.e. (rect. CD, DB) = (sq. on CB). 

Here again the proposition can readily be proved by means of a figure 
similar to that of It. u. 

Draw CA through C at right angles to CB and of length equal to CA in 
the original figure ; make CD double of CA ; 

produce AC to R so that CR= CB. 

Complete the squares on CB and CD, and 
join AD. 

Now we are given the fact that 

(sq. on AS) = 5 (sq. on CA). 
But 
5 (sq. on AC) = (sq. on AC)+ (sq. on CD) 
- (sq. on AD). 
Therefore 

(sq, on AR) = (sq. on AD), 
or AR = AD. 

Now 

(rect KR, RC) + {sq. onAC) = {sq. on AR) 

- (sq, on AD) 

= (sq. on AC) + (sq. on CD). 
Therefore (rect. KR . RC) = (sq. on CD). 

That is, (rectangle RE) = (square CG). 

Subtract the common part CE, 
and (rect. BG) = (sq. RB), 

or rect, CD, DB = (sq. on CB). 

Heiberg, with reason, doubts the genuineness of the Lemma following this 
proposition. 










Proposition 3. 



If a straight line be cut in extreme and mean ratio, the 
square on the lesser segment added to the half of the greater 
segment is five times the square on the half of the greater 
segment. 



44* 



BOOK XIII 



[xm. 3 




For let any straight line AB be cut in extreme and mean 
ratio at the point C, 
let AC be the greater segment, 
and let AC be bisected at D ; 
I say that the square on BD is 
five times the square on DC. 

For let the square AE be 
described on AB, 
and let the figure be drawn 
double. 

Since AC is double of DC, 
therefore the square on AC is 
quadruple of the square on DC, 
that is, RS is quadruple of FG. 

And, since the rectangle AB, BC is equal to the square 
on AC, 

and CE is the rectangle AB, BC, 
therefore CE is equal to RS. 

But RS is quadruple of FG ; 
therefore CE is also quadruple of FG. 

Again, since AD is equal to DC, 
HK is also equal to KF. 

Hence the square GF is also equal to the square HL, 

Therefore GK is equal to KL, that is, MN to NE ; 
hence MF is also equal to FE. 

But MF is equal to CG ; 
therefore CG is also equal to FE. 

Let CN be added to each ; 
therefore the gnomon OPQ is equal to CE. 

But CE was proved quadruple of GF; 
therefore the gnomon OPQ is also quadruple of the square FG. 

Therefore the gnomon OPQ and the square FG are 
five times FG. 

But the gnomon OPQ and the square FG are the 
square DN, 

And DN is the square on DB, and 67*" the square on DC. 

Therefore the square on DB is five times the square 
on DC. q. e. d. 






xiii. 3, 4] 



PROPOSITIONS 3, 4 



447 



In this case we have 

(sq. on SD) = (sq. FG) + (rect. CG) + (rect. CN) 
= (sq. FG) + (rect. J^) + (rect. CN) 
a (sq. ^G) + (rect. C£) 
= (sq. FG) 4 (rect. ^^, BC) 
= (sq. i^iff) + (sq. on AC), by hypothesis, 
= S (sq. oni>C). 
The theorem is still more obvious if the figure 
of ii. 1 1 be used. Let CF be divided in extreme 
and mean ratio at E, by the method of n. n. 
Then, since 

(rect. AB, BC) + (sq. on CD) 
-- sq. on JiD 
= sqs. on CD, CF, 
(rect AB, BC) = (sq. on CF) 
= (sq. on CA), 
and AB is divided at C in extreme and mean ratio. 
And (sq. on BD) = (sq. on DF) 
- s (sq. on CD). 






c 






E F 










/ 


\ 






Proposition 4. 

//" a straight line be cut in extreme and mean ratio, the 
square on the whole and the square on the lesser segment together 
are triple of the square on the greater segment. 

Let AB be a straight line, 
let it be cut in extreme and mean ratio at C, 
and let AC be the greater segment ; 
I say that the squares on AB, BC are 
triple of the square on CA. 

For let the square ADEB be de- 
scribed on AB, 
and let the figure be drawn. 

Since then AB has been cut in extreme 
and mean ratio at C, 
and A C is the greater segment, 

therefore the rectangle AB, BC is equal to the square on A C, 

[vi. Def. 3, vi. 17] 
And AK is the rectangle AB, BC, and HG the square 
on AC-, 
therefore AK is equal to HG. 




448 



BOOK XIII 



[xui, 4, 5 



And, since AF'is equal to FE, 
let CK be added to each ; 

therefore the whole AK is equal to the whole CE ; 
therefore AK, CE are double of AK. 

But AK, CE are the gnomon LMN and the square CK; 
therefore the gnomon LMN and the square CK are double 
of AK. 

But, further, AK was also proved equal to HG ; 
therefore the gnomon LMN and the squares CK, HG are 
triple of the square HG. 

And the gnomon LMN and the squares CK, HG are 
the whole square AE and CK, which are the squares on 
AB, BC, 
while HG is the square on AC. 

Therefore the squares on AB, BC are triple of the square 
on AC. 

Q. E. D. 
Here, as in the preceding propositions, the results are proved tie novo by 
the method of Book n., without reference to that Book, Otherwise the proof 
might have been shorter. 
For, by It. 7, 

(sq. on AB) + (sq. on BC) = 1 (rect, AB, BC) + (sq. on AC) 
= 3 (sq. on A C), 

Proposition 5. 

Lf a straight line be cut in extreme and mean ratio, and 
there be added to it a straight line equal to the greater segment, 
the whole straight line has been cut in extreme and mean ratio, 
and the original straight line is the greater segment. 

For let the straight line AB be cut in extreme and mean 
ratio at the point C, 
let AC be the greater segment, 
and let AD be equal to AC. 

I say that the straight line 
DB has been cut in extreme and 
mean ratio zxA, and the original 
straight line AB is the greater 
segment. 

For let the square AE be described on AB, 
and let the figure be drawn. 



x 



xiii. s, 6] PROPOSITIONS 4—6 449 

Since AB has been cut in extreme and mean ratio at C, 

therefore the rectangle AB, BC is equal to the square on AC. 

[vi. Def. 3, vi. 17] 

And CE is the rectangle AB, BC. and CH the square 
on AC; 
therefore CE is equal to HC. 

But HE is equal to CE, 
and DH is equal to HC ; 
therefore DH is also equal to HE, 

Therefore the whole DK is equal to the whole AE. 

And DK is the rectangle BD, DA, 
for AD is equal to DL ; 
and AE is the square on AB ; 

therefore the rectangle BD, DA is equal to the square 
on AB. 

Therefore, as DB is to BA, so is BA to AD. fvi. 17] 

And DB is greater than BA ; 
therefore BA is also greater than AD. [v. 14] 

Therefore DB has been cut in extreme and mean ratio at 
A, and AB is the greater segment, 

Q. E. D. 

We have (sq. DH) - (sq. HC) 

= (rect. CE), by hypothesis, 
= (rect. HE). 
Add to each side the rectangle A K, and 

(rect. DK) = (sq. AE), 
or (rect. AO, Z>^) = (sq. on AS). 

The result is of course obvious from 11. 11. 

There is an alternative proof given in P after xm. 6, which depends on 
Book v. 

By hypothesis, BA : AC= AC : CS, 

or, inversely, AC : AB = CB -, AC. 

Compo»enHo, {AB + AC):AB = AB : AC, 

or DB:BA = BA . AD. 

Proposition 6. 

If a rational straight line be cut in extreme and mean ratio, 
each of the segments is the irrational straight line called 
apotome. 



45<> BOOK XIII (xtn, 6 

Let AB be a rational straight line, 
let it be cut in extreme and mean 

ratio at C, o % ? ■ 

and let AC be the greater segment ; 

I say that each of the straight lines AC, CB is the irrational 

straight line called apotome. 

For let BA be produced, and let AD be made half of BA. 

Since then the straight line AB has been cut in extreme 
and mean ratio, 

and to the greater segment AC is added AD which is half 
of AB, 
therefore the square on CD is five times the square on DA. 

[xm. l] 

Therefore the square on CD has to the square on DA the 
ratio which a number has to a number ; 

therefore the square on CD is commensurable with the square 
on DA. [x. 6] 

But the square on DA is rational, 

for DA is rational, being half of AB which is rational ; 

therefore the square on CD is also rational ; [x. Def. 4] 

therefore CD is also rational. 

And, since the square on CD has not to the square on 
DA the ratio which a square number has to a square number, 
therefore CD is incommensurable in length with DA ; [x. 9) 
therefore CD, DA are rational straight lines commensurable 
in square only ; 
therefore A C is an apotome. [x. 73] 

Again, since AB has been cut in extreme and mean ratio, 

and AC is the greater segment, 

therefore the rectangle AB, BC is equal to the square on AC 

[vi. Def. 3, vi. 17] 

Therefore the square on the apotome AC, if applied to 
the rational straight line AB, produces BC as breadth. 

But the square on an apotome, if applied to a rational 
straight line, produces as breadth a first apotome ; [x. 97) 

therefore CB is a first apotome. 



xiii. 6, 7] PROPOSITIONS 6, 7 451 



And CA was also proved to be an apotome. 
Therefore etc. 



Q. E. D 



It seems certain that this proposition is an interpolation. P has it, but the 
copyist (or rather the copyist of its archetype) says that " this theorem is not 
found in most copies of the new recension, but is found in those of the old." 
In the first place, there is a scholium to xitt. 17 in P itself which proves the 
same thing as XIII. 6, and which would therefore have been useless if xiii. 6 
had preceded. Hence, when the scholium was written, this proposition had 
not yet been interpolated. Secondly, P has it before the alternative proof of 
xiii. 5 ; this proof is considered, on general grounds, to be interpolated, and 
it would appear that it must have been a later interpolation (xiti. 6) which 
divorced it from the proposition to which it belonged. Thirdly, there is cause 
for suspicion in the proposition itself, for, while the enunciation states that 
each segment of the straight line is an apotome, the proposition adds that the 
lesser segment is a first apotome. The scholium in P referred to has not this 
blot. What is actually wanted in xiii. 1 7 is the fact that the greater segment 
is an apotome. It is probable that Euclid assumed this fact as evident enough 
from xn 1. 1 without further proof, and that he neither wrote XIII. 6 nor the 
quotation of its enunciation in xiii. 1 7. 



Proposition 7. 

If three angles of an equilateral pentagon, taken either in 
order or not in order, be equal, the pentagon will be equiangular. 

For in the equilateral pentagon ABCDE let, first, three 
angles taken in order, those sxA,B, C, 
be equal to one another ; 
I say that the pentagon ABCDE is 
equiangular. 

For let AC, BE, FD be joined. 

Now, since the two sides CB, BA 
are equal to the two sides BA, AE 
respectively, 

and the angle CBA is equal to the 
angle BAE, 
therefore the base AC is equal to the base BE, 
the triangle ABC is equal to the triangle ABE, 
and the remaining angles will be equal to the remaining angles, 
namely those which the equal sides subtend, [1. 4] 

that is, the angle BCA to the angle BEA, and the angle 
ABE to the angle CAB; 
hence the side AF is also equal to the side BE. [1. 6] 




45* BOOK XIII [xiu. 7 

But the whole ACv/as also proved equal to the whole BE; 
therefore the remainder FC is also equal to the remainder FE. 

But CD is also equal to DE. 

Therefore the two sides FC, CD are equal to the two 
sides FE, ED ; 

and the base FD is common to them ; 
therefore the angle FCD is equal to the angle FED. [i. 8] 

But the angle BCA was also proved equal to the angle 
AEB; 

therefore the whole angle BCD is also equal to the whole 
angle A ED. 

But, by hypothesis, the angle BCD is equal to the angles 
at A, B ; 
therefore the angle A ED is also equal to the angles at A, B. 

Similarly we can prove that the angle CDE is also equal 
to the angles at A, B, C ; 
therefore the pentagon ABCDE is equiangular. 

Next, let the given equal angles not be angles taken in 
order, but let the angles at the points A, C, D be equal ; 
I say that in this case too the pentagon ABCDE is equiangular. 

For let BD be joined. 

Then, since the two sides BA, AE are equal to the two 
sides BC, CD, 

and they contain equal angles, 
therefore the base BE is equal to the base BD, 
the triangle ABE is equal to the triangle BCD, 
and the remaining angles will be equal to the remaining angles, 
namely those which the equal sides subtend ; [i. 4] 

therefore the angle AEB is equal to the angle CDB. 

But the angle BED is also equal to the angle BDE, 
since the side BE is also equal to the side BD. [1. 5] 

'Therefore the whole angle AED is equal to the whole 
angle CDE. 

But the angle CDE is, by hypothesis, equal to the angles 
at A, C; 
therefore the angle AED is also equal to the angles at A, C. 



xhi. 7. 8] PROPOSITIONS 7, 8 453 

For the same reason 
the angle ABC is also equal to the angles at A, C, D. 
Therefore the pentagon ABCDE is equiangular. 

Q. E. D. 
This proposition is required in mil 17. 
The steps of the proof may be shown thus. 

I. Suppose that the angles at A, B, C are all equal. 

Then the isosceles triangles BAE, ABC are equal in all respects ; 
thus BE = AC, lBCA = lBEA, lCAB^lEBA. 

By the last equality, FA = FB, 

so that, since BE = AC, FC= FE. 

The As FED, FCD are now equal in all respects, [1. 8, 4) 

and £ FCD - 4 FED. 

But l ACB = l. ABB, from above, 

whence, by addition, l BCD - L AED. 

Similarly it may be proved that U CDE is also equal to any one of the 
angles at A, B, C. 

II. Suppose the angles at A, C, D to be equal. 

Then the isosceles triangles ABE, CBD are equal in all respects, and 
hence BE • BD (so that l BDE = t BED), 
and l CDB = u AEB. 

By addition of the equal angles, 

l CDE = l DEA. 

Similarly it may be proved that l ABC is also equal to each of the angles 
at A, C,D. 

Proposition 8. 

If in an equilateral and equiangular pentagon straight 
lines subtend two angles taken in order, they cut one another 
in extreme and mean ratio, and their greater segments are equal 
to the side of the pentagon. 

For in the equilateral and equiangular pentagon ABCDE 
let the straight lines AC, BE, cutting 
one another at the point H, subtend 
two angles taken in order, the angles 
at A, B ; 

I say that each of them has been 
cut in extreme and mean ratio at 
the point H, and their greater seg- 
ments are equal to the side of the 
pentagon. 

For let the circle ABCDE be 
circumscribed about the pentagon ABCDE. [iv. 14] 




454 BOOK XIII [xui. 8 

Then, since the two straight lines EA, AB are equal to 
the two AS, BC, 
and they contain equal angles, 
therefore the base BE is equal to the base AC, 
the triangle ABE is equal to the triangle ABC, 
and the remaining angles will be equal to the remaining angles 
respectively, namely those which the equal sides subtend, \i. a\ 

Therefore the angle BA C is equal to the angle ABE ; 
therefore the angle A HE is double of the angle BAH. [i. 3 a] 

But the angle EAC is also double of the angle BAC, 
inasmuch as the circumference EDC is also double of the 
circumference CB ; [in. 38, vi. $$] 

therefore the angle HAE is equal to the angle A HE ; 
hence the straight line HE is also equal to EA, that is, to AB. 

And, since the straight line BA is equal to AE, 
the angle ABE is also equal to the angle AEB. [1. 5] 

But the angle ABE was proved equal to the angle BAH; 
therefore the angle BE A is also equal to the angle BAH, 

And the angle ABE is common to the two triangles ABE 
and ABH; 

therefore the remaining angle BAE is equal to the remaining 
angle A HB; [1.33] 

therefore the triangle ABE is equiangular with the triangle 
ABH; 

therefore, proportionally, as EB is to BA, so is AB to BH. 

[vi. 4] 
But BA is equal to EH ; 

therefore, as BE is to EH, so is EH to HB. 

And BE is greater than EH; 
therefore EH is also greater than HB. [v. 14) 

Therefore BE has been cut in extreme and mean ratio at 
H and the greater segment HE is equal to the side of the 
pentagon. 

Similarly we can prove that AC has also been cut in 
extreme and mean ratio at H, and its greater segment CH 
is equal to the side of the pentagon. 

Q. E. D. 



xm. 8, 9 ] PROPOSITIONS 8. 9 455 

In order to prove this theorem we have to show (1) that the As AEB, 
HAB are similar, and (2) that £H= EA (= AB). 
To prove (2) we have 

As AEB, B AC equa\ in all respects, 
whence EB=AC, 

and lBAC^lABE. 

Therefore i. A HE = ilBAC 

= lEAC, 
so that EH= EA 

= AB. 
To prove {1) we have, in the As AEB, HAB, 
lBAH=l.EBA 
= l AEB, 
and l ABE is common , 

therefore the third & s AHB, EAB are equal, 
and A s AEB, HAB are simitar. 

Now, since these triangles are similar, 

EB;BA = BA: BH, 
or (rect. EB, BH) = (sq. on BA) 

= (sq. on EH), 
so that EB is divided in extreme and mean ratio at H 

Similarly its equal, CA, is divided in extreme and mean ratio at H, 

Proposition 9, 

If ike side of the hexagon and that of the decagon inscribed 
in the same circle be added together \ the whole straight line 
has been cut in extreme and mean ratio, and its greater segment 
is the side of the hexagon. 

Let ABC be a circle ; 

of the figures inscribed in the circle ABC let BC be the side 
of a decagon, CD that of a hexagon, 

and let them be in a straight line ; 

I say that the whole straight line 
BD has been cut in extreme and 
mean ratio, and CD is its greater 
segment. 

For let the centre of the circle, 
the point E, be taken, 

let EB, EC, ED be joined, 

and let BE be carried through to A. o 

Since BC is the side of an equilateral decagon, 




456 BOOK XIII [xiil 9 

therefore the circumference ACB is five times the circum- 
ference BC; 
therefore the circumference A C is quadruple of CB. 

But, as the circumference AC is to CB, so is the angle 
ABC to the angle CEB ; [vi. 33] 

therefore the angle A EC is quadruple of the angle CEB. 

And, since the angle EEC is equal to the angle ECB, [1. 5] 
therefore the angle A EC is double of the angle ECB. [1. 32] 

And, since the straight line EC is equal to CD, 
for each of them is equal to the side of the hexagon inscribed 
in the circle ABC, [iv. 15, Por.] 

the angle CED is also equal to the angle CDE ; [1. 5] 

therefore the angle ECB is double of the angle EDC. [1. 31] 

But the angle A EC was proved double of the angle ECB; 
therefore the angle A EC is quadruple of the angle EDC. 

But the angle AEC was also proved quadruple of the 
angle BEC; 
therefore the angle EDC is equal to the angle BEC. 

But the angle EBD is common to the two triangles BEC 
and BED ; 

therefore the remaining angle BED is also equal to the 
remaining angle ECB ; [1. 3*] 

therefore the triangle EBD is equiangular with the triangle 
EEC. 

Therefore, proportionally, as DB is to BE, so is EB to BC. 

[vi. 4] 
But EB is equal to CD. 
Therefore, as BD is to DC, so is DC to CB. 
And BD is greater than DC ; 

therefore DC is also greater than CB, 

Therefore the straight line BD has been cut in extreme 
and mean ratio, and DC is its greater segment. 

Q. E. D. 

BC is the side of a regular decagon inscribed in the circle ; CD is the 
side of the inscribed regular hexagon, and is therefore equal to the radius BE 
or EC. 

Therefore, in order to prove our theorem, we have only to show that tht 
triangles EBC, DBE arc similar. 









xin. 9. i°] PROPOSITIONS 9, 10 45? 

Since BC is the side of a regular decagon, 

(arc BCA) = 5 (arc BC), 
so that (arc CEA) m 4 (arc BC), 

whence l CEA = 4 l BEC. 

But i.CEA = n.ECB. 

Theretore i.ECB = 2i.BEC fi). 

But, since CD = CE, 

L CDE = l CED, 
so that lECB=2lCDE. 

It follows from (1) that lBEC=lCDE. 
Now, in the As EBC, DBE, 

lBEC=l.BDE, 
and & EBC is common, 

so that i. £C# = i. /?£,#, 

and As EBC, DBE are similar. 

Hence DB : BE = EB ; BC, 

or (rect. X>5, JC) = (sq. on ££) 

= (sq. on CD), 
and Z)2? is divided at C in extreme and mean ratio. 

To find the side of the decagon algebraically in terms of the radius we 
have, if x be the side required, 

v + *)* = *I 









m 

whence x = -(J$ — 1). 

2 






Proposition 10. 

If an equilateral pentagon be inscribed in a circle, the 
square on the side of the pentagon is equal to ike squares on 
the side of the hexagon and on that of the decagon inscribed in 
ike same circle. 

Let ASCDE be a circle, 
and let the equilateral pentagon ABCDE be inscribed in the 
circle ABCDE. 

I say that the square on the side of the pentagon ABCDE 
is equal to the squares on the side of the hexagon and on 
that of the decagon inscribed in the circle ABCDE. 

For let the centre of the circle, the point F, be taken, 
let AF be joined and carried through to the point G, 
let FB be joined, 

let FH be drawn from F perpendicular to AB and be carried 
through to K, 



4S8 BOOK XIII [xiii. 10 

let AK, KB be joined, 

let FL be again drawn from /*" perpendicular to AK, and be 
carried through to M, 
and let KN be joined. 

Since the circumference 
ABCG is equal to the circum- 
ference AEDG, 
and in them ABC is equal to 
AED, 

therefore the remainder, the 
circumference CG, is equal to 
the remainder GD. 

But CD belongs to a pen- 
tagon ; 

therefore CG belongs to a 
decagon. 

And, since FA is equal to FB, 
and FH is perpendicular, 
therefore the angle AFK is also equal to the angle KFB, 

[i. S->-*«] 

Hence the circumference AK is also equal to KB \ [in. 26] 

therefore the circumference AB is double of the circumference 

BK, 

therefore the straight line AK is a side of a decagon. 







For the same reason 
AK is also double of KM, 

Now, since the circumference AB is double of the circum- 
ference BK, 
while the circumference CD is equal to the circumference AB, 

therefore the circumference CD is also double of the circum- 
ference BK. 

But the circumference CD is also double of CG ; 

therefore the circumference CG is equal to the circumference 
BK. 

But BK is double of KM, since KA is so also ; 
therefore CG is also double of KM, 



xin. io] PROPOSITION 10 459 

But, further, the circumference CB is also double of the 
circumference BK, 

for the circumference CB is equal to BA. 

Therefore the whole circumference GB is also double 
of BM; 

hence the angle GFB is also double of the angle BFM. [vi. 33] 
But the angle GFB is also double of the angle FAB, 

for the angle FAB is equal to the angle ABF. 

Therefore the angle BFN is also equal to the angle FAB. 
But the angle ABF is common to the two triangles ABF 

and BFN \ 

therefore the remaining angle AFB is equal to the remaining 
angle BNF; ['• 3*] 

therefore the triangle ABF is equiangular with the triangle 
BFN. 

Therefore, proportionally, as the straight line AB is to BF, 
so is FB to BN ; [vi. 4] 

therefore the rectangle AB, BN is equal to the square on BF. 

[vi. 17] 

Again, since AL is equal to LK, 
while LN is common and at right angles, 
therefore the base KN is equal to the base A N ; [1. 4] 

therefore the angle LKN is also equal to the angle LAN. 

But the angle LAN is equal to the angle KBN ; 
therefore the angle LKN is also equal to the angle KBN. 

And the angle at A is common to the two triangles A KB 
znAAKN. 

Therefore the remaining angle AKB is equal to the 
remaining angle KNA ; [1. 32] 

therefore the triangle KB A is equiangular with the triangle 
KNA. 

Therefore, proportionally, as the straight line BA is to 
AK, so is KA to AN; [vi. 4J 

therefore the rectangle BA, AN is equal to the square on AK. 

[vi. 17] 
But the rectangle AB, BN was also proved equal to the 
square on BF; 



460 BOOK XIII [xiii. 10 

therefore the rectangle AB, BN together with the rectangle 
BA, AN, that is, the square on BA [u. a\ is equal to the 
square on BF together with the square on AK. 

And BA is a side of the pentagon, BF of the hexagon 
[iv. is, Por.], and AK of the decagon. 

Therefore etc. 

<i. E. D, 

ABCDE being a regular pentagon inscribed in a circle, and AG the 
diameter through A, tt follows that 

(arc CG) = (arc GD), 
and CG t GD are sides of an inscribed regular decagon. 

FHK being drawn perpendicular to AB, it follows, by I. 26, that 
L s AFK, BFK are equal, and BK, KA are sides of the regular decagon. 
Similarly it may he proved that, FLM being perpendicular to AK, 

AL - LK, 
and (arc AM) = (arc MK). 

The main facts to prove are that 
(1) the triangles ABF, FBN are similar, and (3) the triangles A BK, AKN 
are simitar. 

(1) 2 (arc Ct?) = (arc CD) 

= (arc AB) 
= 2 (arc BK), 
or (arc CG) = (arc BK) = (arc AK) 

= 1 (arc KM). 
And (arc CB) - 2 (arc BK). 

Therefore, by addition, 

(arc BCG) = 1 (arc BKM). 
Therefore u BFG -it- BFN. 

But lBFG=ilFAB, 

so that L FAB = 4 BFN. 

Hence, in the &&ABF, FBN, 

l FAB = l BFN, 
and l ABF is common ; 

therefore l AFB = l BNF, 

and As ABF, FBNvte. similar. 

(2) Since AL- LK, and the angles at L are right, 

AN= NK, 
and LNKA = i.NAK 

m L KBA. 

Hence, in the As ABK, AKN, 

±A£K=lAKN, 

and l. KAN is common, 

whence the third angles are equal ; 

therefore the triangles ABK, AKN are similar. 



xiii. to, n] PROPOSITIONS 10, n 461 

Now from the similarity of A s A BF, FBN it follows that 
AB :BF=BF:BN, 

or (rect. AB, BN) ■ (sq. on BF). 

And, from the similarity of ABK, AKN, 

BA : AK^ AK : AN, 
or (rect. BA, AN) ■ (sq. on AK). 

Therefore, by addition, 

(rect, AB, BN) + (rect. BA, AN) ■ (sq. on BF) + (sq. on AK), 
that is, (sq. on AB) = (sq. on BF) + (sq. on AK). 

If r be the radius of the circle, we have seen (xiii. 9, note) that 
AA'-^Us-i). 

Therefore (side of pentagon) 1 = r* + - (6 - 2 ^5) 

4 

= 7(*«-*VsX 

4 



so that (side of pentagon) = - n/10 - 2 V5 



Proposition i i. 

#V« a *m& ml&£ las &f diameter rational an equilateral 
pentagon be inscribed, the side of the pentagon is the irrational 
straight line called minor. 

For in the circle ABCDE which has its diameter rational 
let the equilateral pentagon ABCDE be inscribed ; 
1 say that the side of the pentagon is the irrational straight 
line called minor. 

For let the centre of the circle, the point F, be taken, 
let AF, FB be joined and carried through to the points, G, H, 
let AC be joined, 
and let FK be made a fourth part of AF. 

Now AF is rational ; 
therefore FK is also rational. 

But BF is also rational ; 
therefore the whole BK is rational. 

And, since the circumference ACG is equal to the circum- 
ference ADG, 

and in them ABC is equal to A ED, 
therefore the remainder CG is equal to the remainder GD. 



4 



BOOK XIII 



[xnt. it 



And, if we join AD, we conclude that the angles at L 
are right, 
and CD is double of CL. 

For the same reason 
the angles at M are also right, 
and AC is double of CM. 







Since then the angle ALC is equal to the angle AMF, 

and the angle LAC is common to the two triangles ACL 

and AMF, 

therefore the remaining angle ACL is equal to the remaining 

angle MFA ; [i. 32] 

therefore the triangle ACL Is equiangular with the triangle 
AMF; 

therefore, proportionally, as LC is to CA, so is MF to /v?. 

And the doubles of the antecedents may be taken ; 

therefore, as the double of LC is to CA, so is the double of 
MF to FA. 

But, as the double of MF is to FA, so is MF to the half 
oiFA\ 

therefore also, as the double of LC is to CA, so is MF to the 
half of T^W. 

And the halves of the consequents may be taken ; 

therefore, as the double of LC is to the half of CA, so is MF 
to the fourth of FA. 



xiii. 1 1] PROPOSITION ii 463 

And DC is double of LC, CM is half of CA, and FK -a. 
fourth part of FA ; 
therefore, as DC is to CM, so is MF to FK. 

Componendo also, as the sum of DC, CM is to CM, so is 
MM to KF\ [v. 18] 

therefore also, as the square on the sum of DC, CM is to the 
square on CM, so is the square on MK to the square on KF. 

And since, when the straight line subtending two sides of 
the pentagon, as AC, is cut in extreme and mean ratio, the 
greater segment is equal to the side of the pentagon, that is, 
to DC, [xui. 8] 

while the square on the greater segment added to the half 
of the whole is five times the square on the half of the 
whole, [xtu. 1] 

and CM is half of the whole AC, 

therefore the square on DC, CM taken as one straight line is 
five times the square on CM. 

But it was proved that, as the square on DC, CM taken 
as one straight line is to the square on CM, so is the square 
on MK to the square on KF ; 
therefore the square on MK is five times the square on KF. 

But the square on KF is rational, 
for the diameter is rational ; 
therefore the square on MK is also rational ; 
therefore MK is rational 

And, since BF is quadruple of FK, 

therefore BK is five times KF ; 

therefore the square on BK is twenty-five times the square 
on KF. 

But the square on MK is five times the square on KF; 
therefore the square on BK is five times the square on KM; 
therefore the square on BK has not to the square on KM 
the ratio which a square number has to a square number ; 

therefore BK is incommensurable in length with KM. [x. 9] 

And each of them is rational. 

Therefore BK, KM are rational straight lines commen- 
surable in square only. 



464 BOOK XIII [xm. ii 

But, if from a rational straight line there be subtracted a 
rational straight line which is commensurable with the whole 
in square only, the remainder is irrational, namely an apotome; 
therefore MB is an apotome and MK the annex to it. [x. 73] 

I say next that MB is also a fourth apotome. 
Let the square on N be equal to that by which the square 
on BK is greater than the square on KM ; 

therefore the square on BK is greater than the square on KM 
by the square on N. 

And, since KF is commensurable with FB, 
eomponendo also, KB is commensurable with FB. [x. 15] 

But BF is commensurable with BH; 

therefore BK is also commensurable with BH. [x. 12] 

And, since the square on BK is five times the square 
on KM, 

therefore the square on BK has to the square on KM the 
ratio which 5 has to 1. 

Theretore, convertendo, the square on BK has to the square 
on N the ratio which 5 has to 4 [v. 19, Por.], and this is not the 
ratio which a square number has to a square number ; 
therefore BK is incommensurable with N\ [x. 9) 

therefore the square on BK is greater than the square on KM 
by the square on a straight line incommensurable with BK. 

Since then the square on the whole BK is greater than 
the square on the annex KM by the square on a straight line 
incommensurable with BK, 

and the whole BK is commensurable with the rational straight 

line, BH, set out, 

therefore MB is a fourth apotome. [x. Deff. in. 4] 

But the rectangle contained by a rational straight line and 
a fourth apotome is irrational, 

and its square root is irrational, and is called minor. [x. 94] 

But the square on AB is equal to the rectangle HB t BM, 

because, when AH is joined, the triangle ABH is equiangular 
with the triangle ABM, and, as HB is to BA, so is AB 
to BM. 



xiii. i r] PROPOSITION ii 4 6 S 

Therefore the side AB of the pentagon is the irrational 
straight line called minor, 

Q. E. D, 

Here we require certain definitions and propositions of Book x. 

First we require the definition of an apotome [see x. 73], which is a straight 
line of the form (p ~ Jk . p), where p is a " rational " straight line and A is any 
integer or numerical fraction, the square root of which is not integral or 
expressible in integers. The lesser of the straight lines p, N Ik, p is the <tn/itx. 

Next we require the definition of th^ fourth apotome [x. Deff. Ill (after 
x, 84)], which is a straight line of the form (x-y), where x, y (being both 
rational and commi able in square only) are also such that Jx'-j" is 
incommensurable wiu. x, while x is commensurable with a given rational 
straight line p. As shown on X. 88 (note), lY.^ fourth apotomt is of the form 

V vi + A/ 

Lastly the miner (straight line) is the irrational straight line defined in 
x. 76. It is of the form (jc — y), where x, y are incommensurable in stjuart, 
and (x" +/*) is ' rational,' while xy is ' medial.' As shown in the note on 
x. 76, the miner irrational straight line is of the form 

±. /, + _*- --E- A - -— - 
V 3 V J7+& V* V •/!*#' 

The proposition may be put as follows. ABCDE being a regular 
pentagon inscribed in a circle, AG, BH the diameters through A, B meeting 
CD in L and AC in At respectively, FK is made equal to \AF. 

Now, the radius AF(r) being rational, so are FK, BK. 

The arcs CG, GD are equal ; 
hence l s at L are right, and CD -zCL 

Similarly l s at M are right, and AC-sCM. 

We have to prove 

(1) that BM is an apotome, 

(a) that BM is a fourth apotome, 

(3) that BA is a minor irrational straight line. 

Remembering that, if L'A is divided in extreme and mean ratio, the 
greater segment is equal to the side of the pentagon [xm. 8], and that accord 
ingly [xm. 1] (CD + \CA) % = 5 (faCA)", we work towards a proportion con- 
taining the ratio (CD + CM) 1 : CAP, thus. 

The As ACL, AFAf&te equiangular and therefore similar. 

Therefore LC: CA = MF:FA, 

and accordingly iZC: CA = MF: \FA ; 

thus %LC:\CA = MF.\FA, 

or DC:CM=MF:FK; 

whence, componendo, and squaring, 

(DC + CMf : CAP - MK* : KF*. 

But {DC+CAf)* = S CAP; 

therefore MK? - $XF*> 



466 BOOK XIIT [xm. n, 13 

[This means that MK* = -& r 1 , 

or JttT = 3&r.] 

4 
It follows that, ■AT*' being rational, MK*, and therefore MK, is rational. 

(1) To prove that BMte an apotome and jWA* its annex. 
We have BF=aFK; 

therefore BX=$FK, 

BK* = a$FK s 

- $MK\ from above ; 
therefore BK* has not to AfJC the ratio of a square number tn a square 
number ; 
therefore BK, MKare incommensurable in length. 

They are therefore rational and commensurable in square only ; 
accordingly BMfc an apotome. 

[BK 1 = $MK> - ?V, and BK= \r. 

Consequently BK - MK= (-*■- — fY ] 

(2) To prove that BM is a fourth apotome. 
First, since KF, FB are commensurable, 

BK, BF are commensurable, i-e. BK is commensurable with BH, a given 
rational straight line. 

Secondly, if W = BK* - KM\ 

since BK*: KM* = $ : 1, 

it follows that BK 1 : JV 1 = 5 ; 4, 

whence #A", /V are incommensurable. 

Therefore BMis a. fourth apotome. 

(3) To prove that BA is a minor irrational straight line. 

If a fourth apotome form a rectangle with a rational straight line, the side 
of the square equivalent to the rectangle is minor [x. 94]. 

Now BA*^HB.BM, 

HB is rational, and BMis a fourth apotome; 
therefore BA is a minor irrational straight line. 

If this is separated into the difference between two straight lines, we have 
BA= r - sis + 2 VS - r ~ ^5 ~ * VS-] 

Proposition 12. 

If an equilateral triangle be inscribed in a circle, the square 
on the side of the triangle is triple of the square on the radius 
of the circle. 




xiu. a, 13] PROPOSITIONS it— 13 467 

Let ABC be a circle, 
and let the equilateral triangle ABC be inscribed in it ; 
I say that the square on one side of 
the triangle ABC is triple of the square 
on the radius of the circle. 

For let the centre D of the circle 
ABC be taken, 

let AD be joined and carried through 

to E, 

and let BE be joined. 

Then, since the triangle ABC is 
equilateral, 

therefore the circumference BEC is a third part of the circum- 
ference of the circle ABC 

Therefore the circumference BE is a sixth part of the 
circumference of the circle ; 

therefore the straight line BE belongs to a hexagon ; 

therefore it is equal to the radius DE. [iv. 15, Por.] 

And, since AE is double of DE, 

the square on AE is quadruple of the square on ED, that is, 
of the square on BE. 

But the square on AE is equal to the squares on AB, BE; 

[111. 31, 1. 47] 
therefore the squares on AB, BE are quadruple of the square 
on BE. 

Therefore, separando, the square on AB is triple of the 
square on BE. 

But BE is equal to DE ; 

therefore the square on AB is triple of the square on DE. 

Therefore the square on the side of the triangle is triple 
of the square on the radius. 

Q. E. D, 

Proposition 13. 

To construct a pyramid, to comprehend it in a given sphere, 
and to prove that the square on the diameter of the sphere is 
one and a half times the square on the side of the pyramid. 



468 



BOOK XIII 



[xm. 13 



Let the diameter AB of the given sphere be set out, 
and let it be cut at the point C so that AC is double of CB ; 
let the semicircle ADB be described on AB, 
let CD be drawn from the point C at right angles to AB, 

and let DA be joined ; 

let the circle EFG which has its radius equal to DC be 
set out, 

let the equilateral triangle EFG be inscribed in the circle EFG, 

[iv. 1] 

let the centre of the circle, the point H, be taken, [ui. 1] 

let EH, HF, HG be joined ; 

from the point H let HK be set up at right angles to the plane 
of the circle EFG, [it is] 

let HK equal to the straight line AC be cut off from HK, 
and let KE, KF, KG be joined. 





Now, since KH is at right angles to the plane of the 
circle EFG, 

therefore it will also make right angles with all the straight 
lines which meet it and are in the plane of the circle EFG. 

[xi. Def. 3] 

But each of the straight lines HE, HF, HG meets it : 

therefore HK is at right angles to each of the straight lines 
HE, HF, HG. 

And, since AC is equal to HK, and CO to HE, 
and they contain right angles, 
therefore the base DA is equal to the base KE. [1. 4] 



xiii. i j] PROPOSITION 13 469 

For the same reason 

each of the straight lines KF, KG is also equal to DA ; 

therefore the three straight lines KE, KF, KG are equal to 
one another. 

And, since AC is double of CB, 

therefore AB is triple of BC. 

But, as AB is to BC, so is the square on AD to the square 
on DC, as will be proved afterwards. 

Therefore the square on AD is triple of the square on DC. 
But the square on FE is also triple of the square on EH, 

[xm. 13] 
and DC is equal to EH ; 

therefore DA is also equal to EF. 

But DA was proved equal to each of the straight lines 
KE, KF, KG ; 

therefore each of the straight lines EF, FG, GE is also equal 

to each of the straight lines KE, KF, KG ; 

therefore the four triangles EFG, KEF, KFG, KEG are 

equilateral. 

Therefore a pyramid has been constructed out of four 
equilateral triangles, the triangle EFG being its base and the 
point K its vertex. 

It is next required to comprehend it in the given sphere 
and to prove that the square on the diameter of the sphere 
is one and a half times the square on the side of the pyramid. 

For let the straight line HL be produced in a straight 
line with KH, 
and let HL be made equal to CB. 

Now, since, as AC is to CD, so is CD to CB, [vi. 8, Por] 

while AC is equal to KH, CD to HE, and CB to HL, 

therefore, as KH is to HE, so is EH to HL ; 

therefore the rectangle KH, HL is equal to the square on 
EH. [vi. 17] 

And each of the angles KHE. EHL is right ; 

therefore the semicircle described on KL will pass through 
E also. [cf. vi. 8, 111. 31.] 



47° 



BOOK XIII 



[xtn. i; 



If then, KL remaining fixed, the semicircle be carried round 
and restored to the same position from which it began to be 
moved, it will also pass through the points F, G, 

since, if FL, LG be joined, the angles at F, G similarly become 
right angles } 

and the pyramid will be comprehended in the given sphere. 

For KL, the diameter of the sphere, rs equal to the 
diameter AB of the given sphere, inasmuch as KH was 
made equal to AC, and HL to CB. 

I say next that the square on the diameter of the sphere 
is one and a half times the square on the side of the 
pyramid 

For, since AC is double of CB, 

therefore AB is triple of BC ; 

and, converiendo, BA is one and a half times AC. 

But, as BA is to AC, so is the square on BA to the square 
on AD. 

Therefore the square on BA is also one and a half times 
the square on AD. 

And BA is the diameter of the given sphere, and AD is 
equal to the side of the pyramid. 

Therefore the square on the diameter of the sphere is 
one and a half times the square on the side of the pyramid. 

Q. E. D. 

Lemma. 

It is to be proved that, as AB is to BC, so is the square 
on AD to the square on DC. 

For let the figure of the semi- 
circle be set out, 
let DB be joined, 
let the square EC be described 
on AC, 

and let the parallelogram FB be 
completed. 

Since then, because the tri- 
angle DAB is equiangular with 
the triangle DAC, 
as BA is to AD, so is DA to AC, 

[vi. 8, vi. 4] 




xiii. 13] PROPOSITION 13 471 

therefore the rectangle BA, AC is equal to the square on AD. 

[vi. 17] 

And since, as AB is to BC, so is EB to BF, [vi. i] 

and EB is the rectangle BA, AC, for EA is equal to AC, 

and BF is the rectangle AC, CB, 

therefore, as AB is to BC, so is the rectangle BA, AC xo the 
rectangle AC, CB. 

And the rectangle BA, AC is equal to the square on AD, 
and the rectangle ^4C, CB to the square on DC, 

for the perpendicular Z?C is a mean proportional between the 
segments AC, CB of the base, because the angle ADB is 
right. [vi. 8, Por.] 

Therefore, as AB is to BC, so is the square on AD to 
the square on DC. 

q. E. D. 

The Lemma is with reason suspected. Euclid commonly takes more 
difficult theorems for granted in the stereometrical Books. It is also clumsy 
in itself, while, from a gloss in the proposition rejected as an interpolation, it 
is clear that the interpolator of the gloss had not the Lemma. With the 
Lemma should disappear the words "as will be proved afterwards " (p. 460), 

In the figure of the proposition, the semicircle really represents half of 
a section of the sphere through its centre and one edge of the inscribed 
tetrahedron (AD being the length of that edge). 

The proof is in three parts, the object of which is to prove 

(1) that KEFG is a tetrahedron with all its edges equal to AD, 

(2) that it is inscribable in a sphere of diameter equal to AB, 

(3) that AB* = %AD*. 
To prove {1) we have to show 

(a) that KE = KF= KG = AD, 

(b) that AD = EF. 

(a) Since HE = HF=HG=CD, 

KH=AC, 
and usACD, KHE, KHF, KHG are right, 

As A CD, KHE, KHF, KHG are equal in all respects ; 






therefore 


KE = KF= KG = AD. 


(6) Since 


AB=$BC, 


and 


AB:BC=AB.AC:AC.CB 




- AD* : CD\ 


it follows that 


AD' = 3 CD\ 


But [xiii, 12] 


EF i =zEH i , 



and EH= CD, by construction 



47^ BOOK XIII [xin, 13 

Therefore AD^EF. 

Thus EFGK'v, a regular tetrahedron. 

(2) We now observe the usefulness of Euclid's description of a sphere 

[in Xt. Def. re- 
producing KJf{= AC) to L so that HL = CB, 

we have KL equal to AB ; 

thus KL is a diameter of the sphere which should circumscribe out tetra- 
hedron, 

and we have only to prove that E, F, G lie on semicircles described on KL 

as diameter. 

E.g. for the point £, 

since AC-.CD^CD.CB, 

while AC=KH, CD^HE, CB = HL, 

we have KH:HE=HE. HL, 

or KH.HL=HE', 

whence, the angles KHE, EHL being right, 

EKL is a triangle right-angled at E [cf. vi, 8]. 

Hence E lies on a semicircle on KL as diameter. 

Similarly for F, G. 

Thus a semicircle on KL as diameter revolving round KL passes 

successively through E, F, G. 

(3) AB = iBC, 

therefore BA = \AC. 

And BA:AC = BA> zBA.AC 

= BA*:AD\ 

Therefore BA* = \AD\ 

If r be the radius of the circumscribed sphere, 

(edge of tetrahedron) = -H— . r = f^/6 . r. 

It will be observed that, although in these cases Euclid's construction is 
equivalent to inscribing the particular regular solid in a given sphere, he does 
not actually construct the solid in the sphere but constructs a solid which a 
sphere equal to the given sphere will circumscribe. Pappus, on the other 
hand, in dealing with the same problems, actually constructs the respective 
solids in the given spheres. His method is to find circular sections in the 
given spheres containing a certain number of the angular points of the given 
solids. His solutions are interesting, although they require a knowledge of 
some properties of a sphere which are of course not found in the Elements 
but belonged to treatises such as the Sphaerica of Theodosius. 

Pappus' solution of the problem of Eucl. XIII. 13. 

In order to inscribe a regular pyramid or tetrahedron in a given sphere, 
Pappus (hi. pp. 142 — 144) finds two circular sections equal and parallel to one 
another, each of which contains one of two opposite edges as its diameter. In 
this and the other similar problems he proceeds in the orthodox manner by 




xiii. 13] PROPOSITION 13 473 

analysis and synthesis. The following is a reproduction of his solution of 
this case. 

Analysis. 

Suppose the problem solved, A, B, C, D being the angular points of the 
required pyramid. 

Through A draw EF parallel to CD; this will make equal angles with 
AC, AD; and, since AB does so too, EF 
is perpendicular to AB [Pappus has a lemma 
for this, p. 140, ii — 34], and is therefore a 
tangent to the sphere (for EF is parallel to 
CD, the base of the triangle A CD, and 
therefore touches the circle circumscribing 
it, while it also touches the circular section 
AB made by the plane passing through AB 
and EF perpendicular to it). 

Similarly GH drawn through D parallel 
to AB touches the sphere. 

And the plane through GH, CD makes 
a circular section equal and parallel to AB. 

Through the centre K of that circular 
section, and in the plane of the section, draw LM perpendicular to CD and 
therefore parallel to AB. Join BL, BM. 

BM\% then perpendicular to AB, LM, and LB is a diameter of the sphere. 

Join MC. 

Then LAP=iMC\ 

and BC=AB = LM, 

so that BC* = 2MC. 

And BM, being perpendicular to the plane of the circle LAf, is perpen- 
dicular to CM, 

whence BC 1 = BM 1 + MC, 

so that BM= MC. 

But BC=ZM; 

therefore LM* = zBM" 1 . 

And, since the angle LMB is right, 

BL>=LM* + MB» = % LM\ 

Synthesis. 

Draw two parallel circular sections of the sphere with diameter a", 
such that 

where d is the diameter of the sphere. 

[This is easily done by dividing BL, any diameter of the sphere, at P, so 
that LP= 2PB, and then drawing PM at right angles to LB meeting the 
great circle LMB of the sphere in M. Then LM « : LB* - LP: LB ' = 2 : 3.] 

Draw sections through M, B perpendicular to MB, and in these sections 
respectively draw the parallel diameters LM, AB. 

Lastly, in the section LM draw CD through the centre K perpendicular 
to LM. 

ABCD is then the required regular pyramid or tetrahedron. 



474 



BOOK XIII 



fxiu, 14 









Proposition 14. 






To construct an octahedron and comprehend it in a sphere, 
as in the preceding case ; and to prove that the square on the 
diameter of the sphere is double of the square on the side of the 

octahedron. 

. 

Lei the diameter AB of the given sphere be set out, 

and let it be bisected at C ; 

let the semicircle ADB be described on AB, 

let CD be drawn from C at right angles to AB, 

let DB be joined ; 

let the square EFGH, having each of its sides equal to DB, 
be set out, 

let HE, EG be joined, 

from the point K let the straight line KL be set up at rig In 
angles to the plane of the square EFGH [xi. 12], and let it be 
carried through to the other side of the plane, as KM ; 

from the straight lines KL, KM let KL, KM be respectively 
cut off equal to one of the straight lines EK, FK, GK, HK, 

and let LE, LF, LG, LH, ME, MF, MG, MM be joined. 




Then, since KE is equal to KH, 
and the angle EKH is right, 
therefore the square on HE is double of the square on EK. 

['■ 47] 
Again, since LK is equal to KE, 

and the angle LKE is right, 

therefore the square on EL is double of the square on EK. 

[id.] 



xiii. 14] PROPOSITION 14 475 

But the square on HE was also proved double of the 
square on EK; 

therefore the square on LE is equal to the square on EH; 
therefore LE is equal to EH. 

For the same reason 
LH is also equal to HE ; 
therefore the triangle LEH is equilateral. 

Similarly we can prove that each of the remaining tri- 
angles of which the sides of the square EFGH are the bases, 
and the points L, M the vertices, is equilateral ; 
therefore an octahedron has been constructed which is con- 
tained by eight equilateral triangles. 

It is next required to comprehend it in the given sphere, 
and to prove that the square on the diameter of the sphere is 
double of the square on the side of the octahedron. 

For, since the three straight lines LK, KM, KE are equal 
to one another, 

therefore the semicircle described on LM will also pass 
through E. 

And for the same reason, 

if, LM remaining fixed, the semicircle be carried round and 
restored to the same position from which it began to be 
moved, 

it will also pass through the points F, G, H, 

and the octahedron will have been comprehended in a sphere. 

I say next that it is also comprehended in the given sphere. 
For, since LK is equal to KM, 

while KE is common, 

and they contain right angles, 

therefore the base LE is equal to the base EM. [1. 4] 

And, since the angle LEM is right, for it is in a semicircle, 

[in. 31J 
therefore the square on LM is double of the square on LE. 

['- 47] 
Again, since AC is equal to CB, 
A Bis double of BC. 



476 



BOOK XIII 



fxm. 14 



But, as AB is to BC, so is the square on AB to the square 
on BD ; 
therefore the square on AB is double of the square on BD. 

But the square on LM was also proved double of the 
square on LE. 

And the square on DB is equal to the square on LE, for 
EH was made equal to DB, 

Therefore the square on AB is also equal to the square 
on LM; 
therefore AB is equal to LM. 

And AB is the diameter of the given sphere ; 
therefore LM is equal to the diameter of the given sphere. 

Therefore the octahedron has been comprehended in the 
given sphere, and it has been demonstrated at the same time 
that the square on the diameter of the sphere is double of the 
square on the side of the octahedron. 

Q. E. D. 

I think [he accompanying figure will perhaps be clearer than that in 
Euclid's text. 

EFGH being a square with side equal to BD, it follows that K£, KF, 
KG, KH are all equal to CB. 





So are KL, KM, by construction ; 
hence LE, LF, LG, i/fand ME, MF, MG, MHslk all equal to EF or BD. 

Thus (1) the figure is made up of eight equilateral triangles and is therefore 
a regular octahedron. 

(a) Since KE^KL = KM, 

the semicircle on LM in the plane LKE passes through E. 

Similarly F, G, -tf lie on semicircles on LAfas diameter. 

Thus all the vertices of the tetrahedron lie on the sphere of which LM is 
a diameter. 

(3) LE = EM=BD; 

therefore LM 1 = iED = t£D' 



or 



= AB\ 
LM=AB. 




xhi. i 4 ] PROPOSITION 14 477 

(4) AS* m iBD i 

= 2£F'. 

If r be the radius of the circumscribed sphere, 

(edge of octahedron) = J 2 . r. 

Pappus' method. 

Pappus (in. pp. 143 — 150) finds the two equal and parallel sections of the 
sphere which circumscribe two opposite faces of the octahedron thus. 

Analysis. 

Suppose the octahedron inscribed, A, B, C; D, E, F being the vertices. 

Through ABC, DEF describe planes 
making the circular sections ABC, DEF. 

Since the straight lines DA, DB, DE, DF 
are equal, the points A, E, F, B lie on a circle 
of which D is the pole. 

Again, since AB, BF, FE, EA are equal, 
A BEE is a square inscribed in the said circle, 
and AB, EF&re parallel. 

Similarly DE is parallel to BC, and DE 
to AC. 

Therefore the circles through D, E, .Fand 
A, B, C are parallel ; and they are also equal 
because the equilateral triangles inscribed in 
them are equal. 

Now, ABC, DEF being equal and parallel circular sections, and AB, EF 
equal and parallel chords " not on the same side of the centres," 
^^"is a diameter of the sphere. 

[Pappus has a lemma for this, pp. 136 — 138], 

And AE = EF, so that AF 1 = tFE\ 

But, if a" be the diameter of the circle DEF, 

d' , = ^EF t . [cf. xhi. 12] 

Therefore, if d be the diameter of the sphere, 

<f*:<r=-3:*- 

Now d is given, and therefore d' is given ; hence the circles DEF, ABC 
are given. 

Synthesis. 

Draw two equal and parallel circular sections with diameter d', such that 

where d is the diameter of the sphere. 

Inscribe an equilateral triangle ABC in either circle {ABC). 

In the other circle draw EF equal and parallel to AB but on the opposite 
side of the centre, and complete the inscribed equilateral triangle DEF. 

ABCDEF'w the octahedron required. 

It will be observed that, whereas in the problem of xnt, 13 Euclid first 
finds the circle circumscribing a face and Pappus first finds an edge, in this 
problem Euclid finds the edge first and Pappus the circle circumscribing 
a face. 



4?8 



BOOK XIII 



[xiii. is 



Proposition 15. 

To construct a cube and comprehend it in a sphere, like the 
pyramid ; and to prove that the square on the diameter of the 
sphere is triple of the square on the side of the cube. 

Let the diameter AB of the given sphere be set out, 
and let it be cut at C so that AC is double of CB ; 
let the semicircle ADB be described on AB, 
let CD be drawn from C at right angles to AB, 
and let DB be joined ; 

let the square EFGH having its side equal to DB be set out, 
from E, F, G, H let EK, FL, GM, HN be drawn at right 
angles to the plane of the square EFGH, 
from EK, FL, GM, HN let EK, FL, GM, HN respectively 
be cut off equal to one of the straight lines EF, EG, 
GH, HE, 

and let KL, LM, MN, NK be joined ; 
therefore the cube FN has been constructed which is contained 
by six equal squares. 

E H . 





It is then required to comprehend it in the given sphere, 
and to prove that the square on the diameter of the sphere is 
triple of the square on the side of the cube. 

For let KG, EG be joined. 

Then, since the angle KEG is right, because KE is also 
at right angles to the plane EG and of course to the straight 
line EG also, [xi. Def. 3] 

therefore the semicircle described on KG will also pass through 
the point E. 

Again, since GF is at right angles to each of the straight 
lines FL, FE, 

GF is also at right angles to the plane FK ; 

hence also, if we join FK, GF will be at right angles to FK ; 



xiii. is] PROPOSITION 15 479 

and for this reason again the semicircle described on GK will 
also pass through F. 

Similarly it will also pass through the remaining angular 
points of the cube. 

If then, KG remaining fixed, the semicircle be carried 
round and restored to the same position from which It began 
to be moved, 
the cube will be comprehended in a sphere. 

I say next that it is also comprehended in the given 
sphere. 

For, since GF is equal to FE, 
and the angle at F is right, 
therefore the square on EG is double of the square on EF. 

But EF is equal to EK; 
therefore the square on EG is double of the square on EK; 
hence the squares on GE, EK, that is the square on GATj.47], 
is triple of the square on EK 

And, since AB is triple of BC, 
while, as AB Is to BC, so is the square on A B to the square 
on BD. 
therefore the square on AB is triple of the square on BD. 

But the square on GK was also proved triple of the square 
onKE. 

And KE was made equal to DB ; 
therefore KG is also equal to AB. 

And AB is the diameter of the given sphere ; 
therefore KG is also equal to the diameter of the given 
sphere. 

Therefore the cube has been comprehended in the given 
sphere ; and it has been demonstrated at the same time that 
the square on the diameter of the sphere is triple of the square 
on the side of the cube. 

Q. 6. a 

AB is divided so that AC = iCB; CD is drawn at right angles to AB, 
and BD is joined 

KG is, by construction, a cube of side equal to BD. 

To prove ( 1) that it is inscribable in a sphere. 

Since KE is perpendicular to EH, EF, 
KE is perpendicular to EG. 



4 8o 



BOOK XIII 



[xm. is 



m 



Thus, KEG being a right angle, £ lies on a semicircle with diameter KG. 
The same thing is proved in the same way of the other vertices 
F, N, L, M, N. 

Thus the cube is inscribed in the sphere of which KG is a diameter, 

KG* = KE' + EG* 
= KE* + 2EF* 
= 3EK*. 
AB=$BC, 
AB.BC= AB* -AB.BC 
= AB 1 :BJ? 1 ; 
AB* = $BD\ 
BD=EK; 
KG = AB. 
AB' = 3BD' 
= SKE'. 
If r be the radius of the circumscribed sphere, 

(edge of cube) ■ A~ .r«f</$. r. 



Also 
while 

therefore 

But 
therefore 

(3) 






<J3 






Pappus' solution. 

In this case too Pappus (ill. pp. 144 — 148) gives the full analysis and 
synthesis. 

Analysis. 

Suppose the problem solved, and let the vertices of the cube be 
A, B, C, A £, F, G, H. 

Draw planes through A, B, C, D and 
E, F t G, H respectively ; these will produce 
parallel circular sections, which are also equal 
since the inscribed squares are equal. 

And CE will be a diameter of the sphere. 

Join EG. 

Now, since EG* = 2EB* - iGC, 
and the angle CGE is right, 

CE" m GC + EG' = %EG\ 

But CE* is given ; 
therefore EG' is given, so that the circles 
EFGH, ABCD, and the squares inscribed in them, are given. 

Synthtsis. 

Draw two parallel circular sections with equal diameters d\ such that 

where d is the diameter of the given sphere. 

Inscribe a square in one of the circles, as ABCD. 

In the other circle draw FG equal and parallel to BC, and complete the 
square on FG inscribed in the circle EFGH. 

The eight vertices of the required cube are thus determined. 




xui. r6] 



PROPOSITIONS 15, 16 



481 






Proposition 16. 



To construct an icosakedron and comprehend it in a sphere, 
tike the aforesaid figures ; and to prove that the side of the 
icosakedron is the irrational straight line called minor. 

Let the diameter AB of the given sphere be set out, 

and let it be cut at C so that AC is quadruple of CB, 

let the semicircle ADB be described on AB, 

let the straight line CD be drawn from C at right angles 
to AB, 

and let DB be joined ; 




let the circle EFGHK be set out and let its radius be equal 

toX^, 

let the equilateral and equiangular pentagon EFGHK be 

inscribed in the circle EFGHK, 

let the circumferences EF, FG, GH, HK, KE be bisected at 

the points L, M, N, O, P, 

and let LM, MN, NO, OP, PL, EP be joined. 



4 8a BOOK XIII [xhi. 16 

Therefore the pentagon LMNOP is also equilateral, 
and the straight line EP belongs to a decagon. 

Now from the points E, F, G, H, K\et the straight lines 
EQ, FR, GS, HT, KU be set up at right angles to the plane 
of the circle, and let them be equal to the radius of the circle 
EFGHK, 

let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, N'T, 
TO, OU, UP, PQ be joined. 

Now, since each of the straight lines EQ, KU is at right 
angles to the same plane, 
therefore EQ is parallel to KU. [xi. 6] 

But it is also equal to it ; 
and the straight lines joining those extremities of equal and 
parallel straight lines which are in the same direction are equal 
and parallel. ['• 33] 

Therefore Q U is equal and parallel to EK, 

But EK belongs to an equilateral pentagon ; 
therefore £?f/a!so belongs to the equilateral pentagon inscribed 
in the circle EFGHK. 

For the same reason 
each of the straight lines QR, RS, ST, TU also belongs to 
the equilateral pentagon inscribed in the circle EFGHK \ 
therefore the pentagon QRSTU is equilateral. 

And, since QE belongs to a hexagon, 
and EP to a decagon, 
and the angle QEP is right, 
therefore QP belongs to a pentagon ; 

for the square on the side of the pentagon is equal to the 
square on the side of the hexagon and the square on the side 
of the decagon inscribed in the same circle. [xm. 10] 

For the same reason 
PU is also a side of a pentagon. 

But QU also belongs to a pentagon ; 
therefore the triangle QP U is equilateral. 

For the same reason 
each of the triangles QLR, RMS, SNT, TOU'is also equi- 
lateral. 



xni. i6] PROPOSITION t6 483 

And, since each of the straight lines QL, QP was proved 
to belong to a pentagon, 
and LP also belongs to a pentagon, 
therefore the triangle QLP is equilateral. 

For the same reason 
each of the triangles LRM, MSN, NTO, O UP is also equi- 
lateral. 

Let the centre of the circle EFGHK. the point V, be 
taken ; 

from V let VZ be set up at right angles to the plane of the 
circle, 

let it be produced in the other direction, as VX, 
let there be cut off VW, the side of a hexagon, and each of 
the straight lines VX, WZ, being sides of a decagon, 
and let QZ, QW, UZ, EV, L V, LX, XM be joined. 

Now, since each of the straight lines VW, QE is at right 
angles to the plane of the circle, 
therefore VWis parallel to QE, [xi. 6] 

But they are also equal ; 
therefore EV, Q W are also equal and parallel. [1. 33] 

But E V belongs to a hexagon ; 
therefore QW also belongs to a hexagon. 

And, since Q W belongs to a hexagon, 
and WZ to a decagon, 
and the angle Q WZ is right, 
therefore QZ belongs to a pentagon. [xm. 10] 

For the same reason 
UZ also belongs to a pentagon, 

inasmuch as, if we join VK, WU, they will be equal and 
opposite, and VK, being a radius, belongs to a hexagon ; 

[iv. 15, Por] 
therefore WU also belongs to a hexagon. 

But WZ belongs to a decagon, 
and the angle U WZ is right ; 
therefore UZ belongs to a pentagon. [xm. 10] 

But Q U also belongs to a pentagon ; 
therefore the triangle QUZ is equilateral. 



484 BOOK XIII [xm. id 

For the same reason 
each of the remaining triangles of which the straight lines 
QR, RS, ST, 7"£/are the bases, and the point Z the vertex, 
is also equilateral. 

Again, since VL belongs to a hexagon, 
and VX to a decagon, 
and the angle L VX is right, 
therefore LX belongs to a pentagon. [xm. w] 

For the same reason, 
if we join MV, which belongs to a hexagon, 
MX is also inferred to belong to a pentagon. 

But LM also belongs to a pentagon ; 
therefore the triangle LMX is equilateral. 

Similarly it can be proved that each of the remaining 
triangles of which MN, NO, OP, PL are the bases, and the 
point X the vertex, is also equilateral. 

Therefore an icosahedron has been constructed which is 
contained by twenty equilateral triangles. 

It is next required to comprehend it in the given sphere, 
and to prove that the side of the icosahedron is the irrational 
straight line called minor. 

For, since VW belongs to a hexagon, 
and WZ to a decagon, 

therefore VZ has been cut in extreme and mean ratio at W, 
and VW is its greater segment ; [xm 9] 

therefore, as ZV'vs, to VW, so is VWto WZ. 

But V Wis equal to VE, and WZ to VX; 
therefore, as ZVis to VE, so is EVto VX. 

And the angles ZVE, EVX are right ; 
therefore, if we join the straight line EZ, the angle XEZ 
will be right because of the similarity of the triangles XEZ, 
VEZ. 

For the same reason, 
since, as ZVis to VW, so is f-^to WZ, 
and ZV\s equal to XW, and VWto WQ, 
therefore, as XW is to WQ, so is QW to WZ. 



xur. 16] PROPOSITION 16 485 

And for this reason again, 
if we join QX, the angle at Q will be right ; [vi. 8] 

therefore the semicircle described on XZ will also pass 
through Q. [in. 31] 

And if, XZ remaining fixed, the semicircle be carried 
round and restored to the same position from which it began 
to be moved, it will also pass through Q and the remaining 
angular points of the icosahedron, 

and the icosahedron will have been comprehended in a 
sphere. 

I say next that it is also comprehended in the given sphere. 
For let VW be bisected at A'. 

Then, since the straight line VZ has been cut in extreme 
and mean ratio at W, 

and ZW is its lesser segment, 

therefore the square on ZW added to the half of the greater 
segment, that is WA', is five times the square on the half 
of the greater segment ; [xm. 3] 

therefore the square on ZA' is five times the square on 
A'W. 

And ZX is double of ZA\ and VW double of A'W; 

therefore the square on ZX is five times the square on 

wv. 

And, since AC is quadruple of CB, 

therefore AB is five times BC. 

But, as A B is to BC, so is the square on AB to the square 
on BD ; [vi. 8, v. Def. 9 ] 

therefore the square on AB is five times the square on BD, 

But the square on ZX was also proved to be five times 
the square on VW. 

And DB is equal to VW, 

for each of them is equal to the radius of the circle EFGHK\ 

therefore AB is also equal to XZ. 

And AB is the diameter of the given sphere ; 
therefore XZ is also equal to the diameter of the given sphere. 

Therefore the icosahedron has been comprehended in the 
given sphere 



4 86 BOOK XIII (xili. 16 

I say next that the side of the icosahedron is the irrational 
straight line called minor. 

For, since the diameter of the sphere is rational, 
and the square on it is five times the square on the radius 
of the circle EFGHK, 

therefore the radius of the circle EFGHK is also rational ; 
hence its diameter is also rational. 

But, if an equilateral pentagon be inscribed in a circle 
which has its diameter rational, the side of the pentagon is 
the irrational straight line called minor. [xm. 11] 

And the side of the pentagon EFGHK is the side of the 
icosahedron. 

Therefore the side of the icosahedron is the irrational 
straight line called minor. 

Porism. From this it is manifest that the square on the 
diameter of the sphere is five times the square on the radius 
of the circle from which the icosahedron has been described, 
and that the diameter of the sphere is composed of the side 
of the hexagon and two of the sides of the decagon inscribed 
in the same circle. 

Q. E. D. 

Euclid's method is 
(1) to find the pentagons in the two parallel circular sections of the sphere, 
the sides of which form ten (five in each circle) of the edges of the icosahedron, 
(2} to find the two points which are the poles of the two circular sections, 

(3) to prove that the triangles formed by joining the angular points of the 
pentagons which are nearest to one another two and two are equilateral, 

(4) to prove that the triangles of which the poles are the vertices and the 
sides of the pentagons the bases are also equilateral, 

(5) that all Che angular points Other than the poles lie on a sphere the 
diameter of which is the straight line joining the poles, 

(6) that this sphere is of the same size as the given sphere, 

(7) that, if the diameter of the sphere is rational, the edge of the icosahedron 
is the minor irrational straight line. 

I have drawn another figure which will perhaps show the pentagons, and 
the position of the poles with regard to them, more clearly than does Euclid's 
figure. 

(1) If A3 is the diameter of the given sphere,, divide AB at C so that 

AC=a,CB; 
draw CD at right angles to AB meeting the semicircle on AB in D. 
Join BD. 



xiu. i6] 



PROPOSITION 1 6 



487 



BD is the radius of the circular sections containing the pentagons. 

[If r is the radius of the sphere, 
since AB :BC=AB*:AB .BC 

= AB':BD\ 
while AB=sBC, 

it follows that AB* = $3D\ 

or (radius of section)' • \r*. 

f* 

Thus [xiu. 10, note] (side of pentagon)' = — (10- tJs)-] 











Inscribe the regular pentagon EFGHK in the circle EFGHK of radius 
equal to BD. 

Bisect the arcs EF, FG, . . ., so forming a decagon in the circle. 

Joining successive points of bisection, we obtain another regular pentagon 
LMNOP. 

LMNOP is one of the pentagons containing five edges of the icasahedron. 

The other circle and inscribed pentagon are obtained by drawing perpen- 
diculars from E, F, G, H, K to the plane of the circle, as EQ, FB, GS, 
HT, KU, and making each of these perpendiculars equal to the radius of the 
circle, or, as Euclid says, the side of the regular hexagon in it. 

QRSTU is the second pentagon (of course equal to the first) containing five, 
edges of the icosahedron. 

Joining each angular point of one of the two pentagons to the two nearest 
angular points in the other pentagon, we complete ten triangles each of which 
has for one side a side of one or other of the two pentagons. 

V, W are the centres of the two circles, and VW is of course perpen- 
dicular to the planes of both. 

(2) Produce VW in both directions, making VX and IVZ both equal to 
a side of the regular decagon in the circle (as EL). 

Joining X, Z to the angular points of the corresponding pentagons, we 



4 88 BOOK XIII [xih. 16 

have five more triangles formed with the sides of each pentagon as bases, ten 
more triangles in all. 

Now we come to the proof. 
{3) Taking two adjacent perpendiculars, EQ, KU, to the plane of the circle 
EFGHK, we see that they are parallel as well as equal; 
therefore QU, EK are equal and parallel. 
Similarly for QR, EPetc. 
Thus the pentagons have their sides equal. 

To prove that the triangles QPL etc., are equilateral, we have, e.g. 
QL* = ££* + £()* 

= (side of decagon) 1 + (side of hexagon)* 
= (side of pentagon) 5 , [xill. 10] 

Le, QL - ^side of pentagon in circle) 

= LP. 
Similarly QP = LP, 

and A QPL is equilateral. 

So for the other triangles between the two pentagons. 
(4) Since VW, EQ are equal and parallel, 

VE, WQ are equal and parallel. 
Thus WQ is equal to the side of a regular hexagon in the circles. 
Now the angle ZWQ is right ; 
therefore ZQ = ZW* + WQ 1 

= (side of decagon)* + (side of hexagon)* 
= (side of pentagon)*. [xiti. 10] 

Thus ZQ, ZR, ZS, ZT, ZU are all equal to QR, RS etc. ; and the 
triangles with Z as vertex and bases QR, RS etc. are equilateral. 

Similarly for the triangles with A" as vertex and LM, AfN f&c. as bases. 
Hence the figure is an icosahedron, being contained by twenty equal 
equilateral triangles. 

(5) To prove that all the vertices of the icosahedron lie on the sphere 
which has XZ for diameter. 

VW being equal to the side of a regular hexagon, and WZ to the side of 
a regular decagon inscribed in the same circle, 
VZ is divided at W\x\ extreme and mean ratio. [xm. 9] 

Therefore ZV \ VW = VW : WZ, 

or, since VW = VE, WZ=VX, 

ZV: VE= VE: VX. 
Thus £ lies on the semicircle on ZX as diameter. [vi. 8] 

Similarly for all the other vertices of the icosahedron. 
Hence the sphere with diameter XZ circumscribes it, 

(6) To prove XZ.r.AS. 

Since VZ is divided in extreme and mean ratio at W, and VW is 
bisected at A', 

AT=iA'W i . [xm, 3] 

Taking the doubles of A'Z, A' W, we have 

= A&. [see under (1) above] 



xiii. i6] PROPOSITION 16 4»9 

That is, XZ=AB. 

[If r is the radius of the sphere, 

VW=£D=-$-r, 

Vs 

VX = (side of decagon in circle of radius BD) 

SD Us-i) [xm. 9 ,note] 





a 






-£«*-* 




Consequently 


XZ= VW+2VX 






= i"7s rU5 - 


«> 




*«*] 





2 



(6) The radius of the circle EFGHK is equal to -j- r, and is therefore 

" rational " in Euclid's sense. 

Hence the side of the inscribed pentagon is the irrational straight line 
Called minor. [xlll. nj 

[The side of this pentagon is the edge of the icosahedron, and its value is 
(note on xm. 10) 



BD / — —r 
= gVio(5-V5)-] 









Pappus* solution. 

This solution (Pappus, in. pp. 150 — 6) differs considerably from that of 
Euclid. Whereas Euclid uses two circular sections of the sphere (those- 
circumscribing the pentagons of his construction), Pappus finds four parallel 
circular sections each passing through three of the vertices of the icosahedron; 
two of the circles are small circles circumscribing two opposite triangular 
faces respectively, and the other two circles are between these two circles, 
parallel to them and equal to one another. 

Analysis. 

Suppose the problem solved, the vertices of the icosahedron being A, B, C; 
D, E, F; G,H,K,L,M,N. 

Since the straight lines BA, BC, BF, BO, BE drawn from B to the 
surface of the sphere are equal, 

A, C, F, G, E are in one plane. 

And AC, CF, FG, GF, FA are equal ; 

therefore ACFGE is an equilateral and equiangular pentagon. 

So are the figures KEBCD, DHFBA, AKLGB, AKNHC, and 
CHMGB. 

Join EF, KH. 



49° 



BOOK xnr 



[XIII. 1C1 



Now AC will be parallel to EF (in the pentagon ACFGE) and to KH 
(in the pentagon AKNHC), so that EF, KH are also parallel ; 

and further Aiff is parallel to LM (in the pentagon LKDHM). 




Similarly .8C, £Z», GH, ZiVare all parallel; 
and likewise BA, FD, GK, J/Ware all parallel. 

Since BC is equal and parallel to LN t and BA to JtfjV, the circles ABC, 
LMN&k equal and parallel. 

Similarly the circles DEF, KG Hare equal and parallel ; for the triangles 
inscribed in them are equal (since each of the Sides in both is the chord 
subtending an angle of equal pentagons), and their sides are parallel re- 
spectively. 

Now in the equal and parallel circles DEF, KGH the chords EF, KH 
are equal and parallel, and on opposite sides of the centres ; 
therefore FK is a diameter of the sphere [Pappus' lemma, pp. 136 — 8], and the 
angle FEK\% right [Pappus' lemma, p. 138, 20 — 26]. 

[The diameter FKis not actually drawn in the figure.] 

In the pentagon GEACF, if EF be divided in extreme and mean ratio, 

the greater segment is equal to AC. [Eucl. xm, 8] 

Therefore EF : AC= (side of hexagon) ; (side of decagon in same circle). 

[xm. t,] 
And EF 1 + AC 1 = EF> + £/[* = <?, 

where d is the diameter of the sphere. 

Thus FK, EF, AC are as the sides of the pentagon, hexagon and decagon 
respectively inscribed in the same circle. [xm. 10] 

But FK, the diameter of the sphere, is given ; 

therefore EF, AC ate given respectively ; 

thus the radii of the circles EFD, A CB are given (if r, r 1 are their radii, 

r> = $EF*,r> = bAC). 



xin. i6] PROPOSITION 16 491 

Hence the circles are given ' 
and so are the circles KHG, LMN which are equal and parallel to them 
respectively. 

Synthesis. 

If d be the diameter of the sphere, set out two straight lines x, y, such 
that d, x, y are in the ratio of the sides of the pentagon, hexagon and decagon 
respectively inscribed in one and the same circle. 

Draw (r) two equal and parallel circular sections in the sphere, with radii 
equal to r, where f" = J^, as DEF, KGH, 

and (a) two equal and parallel circular sections is ABC, LMN, with radius t* 
such that r* = \y*. 

In the circles (1) draw EF. KHzs sides of inscribed equilateral triangles, 
parallel to one another, and on opposite sides of the centres ; 
and in the circles (z) draw AC, LM&& sides of inscribed equilateral triangles 
parallel to one another and to EF, KH, and so that AC, EFaiz on opposite 
sides of the centres, and likewise KH, LM. 

Complete the figure. 

The correctness of the construction is proved as in the analysis. 

It follows also (says Pappus) that 

(diam. of sphere}* = 3 (side of pentagon in DEF) 1 . 

For, by construction, KF : FE =p : h, 

where/, h are the sides of the pentagon and hexagon inscribed in the same 
circle DEF. 

And FE : h = the ratio of the side of an equilateral triangle to that of a 
hexagon inscribed in the same circle ; 
that is, FE 1 Am Jfj s r, 

whence KF -.p = Ji ■ i, 

or KF*=tf. 

Another construction. 

Mr H. M. Taylor has a neat construction for an icosahedron of edge a. 

Let / be the length of the diagonal of a regular pentagon with side equal 
to a. 

Then (figure of xnt. 8), by Ptolemy's theorem, 
P = ia + a*. 

Construct a cube with edge equal to /. 

Let O be the centre of the cube. 

From O draw OL, OM, ON perpendicular to three adjacent faces, and in 
these draw PF, QQ, RR parallel to AB, AD, AE respectively. 

Make LP, LP', MQ, MQ, NR, NK all equal to Jo. 

Le [ Pi P> 9i 9> r i r ' be the reflexes of P, P', Q, Q, K, R' respectively. 

Then will P, P\ Q, Q, P, K, p, p', g, q\ r, r be the vertices of a regular 
icosahedron. 

The projections of PQ on AB, AD, AE are equal to j (/-«), J a, \l 
rc s dgc ti vcl Vj 

Therefore PCf = i (/ - a)' + J a* + \ /> 

= «*. 



49* 



BOOK XIII 



[xm. 16 



Therefore PQ = a. 

Similarly it may be proved that every other edge is equal to a 

All the angular points tie on a sphere with radius OP, and 

op* = i («* + /•>. 













Each solid pentahedral angle is composed of five equal plane angles, each 
of which is the angle of an equilateral triangle. 
Therefore the icosahedron is regular. 

And, from the equation i*-la + a\ we derive 



Therefore, if r-be the radius of the sphere, 



4»". 



whence 
as above.] 



= 4r ls /io-s i ys/V8o 
= ^ Vio(5-7sJ, 






xm. 17J 



PROPOSITIONS 16, 17 



493 






Proposition r 7. 



To construct a dodecahedron and comprehend, it in a sphere, 
like the aforesaid figures, and to prove that the side of the 
dodecahedron is the irrational straight line called apotome. 

Let ABCD, CBEE, two planes of the aforesaid cube at 
right angles to one another, be set out, 

let the sides AB, BC, CD, DA, EF, EB, EC be bisected at 
G, H, K, L, M, N, respectively, 
let GK, HL, MH, NO be joined, 

let the straight lines NP, PO, HQ be cut in extreme and 
mean ratio at the points R, S, T respectively, 
and let RP, PS, TQ be their greater segments ; 
from the points R, S, 7* let RU, SV, TlVbe set up at right 
angles to the planes of the cube towards the outside of the 
cube, 

let them be made equal to RP, PS, TQ, 

and let UB, BW, JVC, CV, VU be joined. 







I say that the pentagon UBWCV is equilateral, and in 
one plane, and is further equiangular. 
For let RB, SB, VB be joined. 



494 BOOK XIII [xiu. 17 

Then, since the straight line NP has been cut in extreme 
and mean ratio at R, 
and RP is the greater segment, 

therefore the squares on PN, NR are triple of the square 
on RP. [xiu. 4J 

But PN is equal to NB, and PR to RU; 
therefore the squares on BN, NR are triple of the square 
on RU. 

But the square on BR is equal to the squares on BN, NR; 

[I- «] 

therefore the square on BR is triple of the square on RU; 
hence the squares on BR, RU are quadruple of the square 
on RU, 

But the square on B U is equal to the squares on BR, RU; 
therefore the square on BUis quadruple of the square on RU; 
therefore BU Is double of RU. 

But VU is also double of UR, 
inasmuch as SR is also double of PR, that is, of RU; 
therefore BU is equal to UV. 

Similarly it can be proved that each of the straight lines 
BfV, WC, CV is also equal to each of the straight lines 
BU, UV. 

Therefore the pentagon BUVCW\s equilateral. 

I say next that it is also in one plane. 

For let PX be drawn from P parallel to each of the 
straight lines RU, SV and towards the outside of the cube, 
and let XH, HWbe, joined ; 
I say that XHW is a straight line. 

For, since HQ has been cut in extreme and mean ratio at 
T, and Q T is its greater segment, 
therefore, as HQ is to Q T, so is £?7^to TH. 

But HQ is equal to HP, and QT to each of the straight 
lines TW,PX\ 
therefore, as HP is to PX, so is WT to TH. 

And HP is parallel to TW, 
for each of them is at right angles to the plane BD ; fxi. 6] 
and TH is parallel to PX, 
for each of them is at right angles to the plane BF. [**] 



xiii. 17] PROPOSITION 17 495 

But if two triangles, as XPH, HTW, which have two 
sides proportional to two sides be placed together at one 
angle so that their corresponding sides are also parallel, 
the remaining straight lines will be in a straight line ; [vi. $%\ 
therefore XH is in a straight line with HW. 

But every straight line is in one plane ; [xi. 1] 

therefore the pentagon [/BWC Vis in one plane. 

I say next that it is also equiangular. 

For, since the straight line NP has been cut in extreme 
and mean ratio at R, and PR is the greater segment, 
while PR is equal to PS, 

therefore AfS has also been cut in extreme and mean ratio 
at P, 

and NP is the greater segment ; [xiii. 5] 

therefore the squares on NS, SP are triple of the square 
on NP. [xiii. 4] 

But NP is equal to NS, and PS to SV; 
therefore the squares on NS, SV are triple of the square 
on NB; 

hence the squares on VS, SN, NB are quadruple of the square 
on NB. 

But the square on SB is equal to the squares on SN, NB ; 
therefore the squares on BS, SV, that is, the square on BV 
—for the angle VSB is right — is quadruple of the square 
on NB; 

therefore VB is double of BN. 

But BC is also double of BN ; 

therefore BV is equal to BC. 

And, since the two sides BU, UV are equal to the two 
sides BW, WC, 

and the base BV is equal to the base BC, 

therefore the angle BUV is equal to the angle BWC. [1. 8] 

Similarly we can prove that the angle (JVC is also equal 
to the angle BWC; 

therefore the three angles BWC BUV, UVC are equal to 
one another. 



49*5 BOOK XIII [xih. 17 

But if in an equilateral pentagon three angles are equal to 
one another, the pentagon will be equiangular, [xui. 7] 

therefore the pentagon BUVCW\s equiangular. 

And it was also proved equilateral ; 
therefore the pentagon BUVCW is equilateral and equi- 
angular, and it is on one side BC of the cube. 

Therefore, if we make the same construction in the case 
of each of the twelve sides of the cube, 

a solid figure will have been constructed which is contained 
by twelve equilateral and equiangular pentagons, and which is 
called a dodecahedron. 

It is then required to comprehend it in the given sphere, 
and to prove that the side of the dodecahedron is the irrational 
straight line called apotome. 

For let XP be produced, and let the produced straight 
line be XZ ; 

therefore PZ meets the diameter of the cube, and they bisect 
one another, 

for this has been proved in the last theorem but one of the 
eleventh book. [xi. 38] 

Let them cut at Z ; 
therefore Z is the centre of the sphere which comprehends 
the cube, 
and ZP is half of the side of the cube. 

Let UZ be joined. 

Now, since the straight line NS has been cut in extreme 
and mean ratio at P, 
and NP is its greater segment, 

therefore the squares on NS, SP are triple of the square 
on NP. [xui. 4] 

But NS is equal to XZ, 
inasmuch as NP is also equal to PZ, and XP to PS, 

But further PS is also equal to XU, 
since it is also equal to RP ; 

therefore the squares on ZX, XU are triple of the square 
on NP. 

But the square on UZ is equal to the squares on ZX, XU; 
therefore the square on UZ is triple of the square on NP. 



xni. 17] PROPOSITION r7 497 

But the square on the radius of the sphere which compre- 
hends the cube is also triple of the square on the half of the 
side of the cube, 

for it has previously been shown how to construct a cube and 
comprehend it in a sphere, and to prove that the square on 
the diameter of the sphere is triple of the square on the side 
of the cube. [xm, 15] 

But, if whole is so related to whole, so is half to half also ; 
and NP is half of the side of the cube ; 

therefore UZ is equal to the radius of the sphere which com- 
prehends the cube. 

And Z is the centre of the sphere which comprehends the 
CHbe ; 
therefore the point U is on the surface of the sphere. 

Similarly we can prove that each of the remaining angles 
of the dodecahedron is also on the surface of the sphere ; 
therefore the dodecahedron has been comprehended in the 
given sphere. 

I say next that the side of the dodecahedron is the irrational 
straight line called apotome. 

For since, when NP has been cut in extreme and mean 
ratio, RP is the greater segment, 

and, when PO has been cut in extreme and mean ratio, PS 
is the greater segment, 

therefore, when the whole NO is cut in extreme and mean 
ratio, RS is the greater segment. 

[Thus, since, as NP is to PR, so is PR to RN, 
the same is true of the doubles also, 

for parts have the same ratio as their equimultiples ; [v. 15) 
therefore as NO is to RS, so is RS to the sum of NR, SO. 

But NO is greater than RS ; 
therefore RS is also greater than the sum of NR, SO ; 
therefore NO has been cut in extreme and mean ratio, 
and RS is its greater segment.] 

But RS is equal to UV \ 
therefore, when NO is cut in extreme and mean ratio, UV is 
the greater segment. 



498 BOOK XIII [xiii. 17 

And, since the diameter of the sphere is rational, 
and the square on it is triple of the square on the side of the 
cube, 
therefore NO, being a side of the cube, is rational. 

[But if a rational line be cut in extreme and mean ratio, 
each of the segments is an irrational apotome.] 

Therefore UV, being a side of the dodecahedron, is an 
irrational apotome. [xiii. 6] 

Porism. From this it is manifest that, when the side of 
the cube is cut in extreme and mean ratio, the greater segment 
is the side of the dodecahedron. 

Q. E. D. 

In this proposition we find Euclid using two propositions which precede 
but are used nowhere else, notably vi. 32, which some authors, in consequence 
of their having overlooked its use here, have been hard put to it to explain. 

Euclid's construction in this case is really identical with that given by 
Mr H. M, Taylor, and also referred to by Henrici and Treutlein under " crystal- 
formation." 

Euclid starts from the cube inscribed in a sphere, as in xiii. 15, and then 
finds the side of the regular pentagon in which the side of the cube is a 
diagonal 

Mr Taylor takes / to be the diagonal of a regular pentagon of side a, 
so that, by Ptolemy's theorem, 

P - a I + a*, 
constructs a cube of wnich / is the edge, and gets the side of the pentagon 
by drawing ZX from Z, the centre of the cube, perpendicular to the face BF 
and equal to J (/+«), then drawing UV through X parallel to BC, and 
making UX, XV both equal to \a. 

Euclid finds UVtinxt. 

Draw NO, Mil bisecting pairs of opposite sides in the square BF and 
meeting in P. 

Draw GK, HL bisecting pairs of opposite sides in the square BD and 
meeting in Q. 

Divide FN, FO, QH respectively in extreme and mean ratio at R, S, T 
(FR, FS, QT being the greater segments); draw RU, SV, TW outwards 
perpendicular to the respective faces of the cube, and all equal in length 
to FR.FS.TQ. 

Join BU, UV, VC, CW, WB. 

Then BUVCW is one of the pentagonal faces of the dodecahedron ; 
and the Others can be constructed in the same way. 

Euclid now proves 

(1) that the pentagon BUVCW\$ equilateral, 

(1) that it is in one plane, 

(3) that it is equiangular, 



xiii. 17] 



PROPOSITION 17 



499 



(4) that the vertex U is on the sphere which circumscribes the cube, and 
hence 

{5) that all the other vertices lie on the same sphere, 
and (6) that the side of the dodecahedron is an apotome, 

(1) To prove that the pentagon BUVCW'vz equilateral. 
We have BU*= BR 1 + RU* 

= ( BN* + MR*) -h RP* 

- (FN* + NR*) + RP* 

= SUP* * RP* [xui. 4] 

= 4^/" 

= UV*. 
Therefore BU = UV. 













Similarly it may be proved that BW, JVC, CV are all equal to UV 
or BU. 

J Mr Taylor proceeds in this way. With his notation, the projections of 
on BA, BC, BE are respectively Jo, J (/- a), \l. 

Therefore BU* = ia , + J(/-a) , + \l* 

= \(l t -ai+a t ) 

= a*. 
Similarly for BW, SVC etc.] 

(2) To prove that the pentagon BUVCW\& in one plane. 
Draw PX parallel to R Uoi S'V meeting UV in X. 
Join XH, HW. 






5<x> BOOK XIII (xiil t7 

TKen we have to prove that XH, HW are in one straight tine. 

Now HP, WT, being both perpendicular to the face BD, are parallel. 

For the same reason XP, HT are parallel. 

Also, since QH is divided at T in extreme and mean ratio, 
QH:QT=QT:Tff. 

And QH= HP, QT= WT= PX. 

Therefore HP .PX^WT. TH. 

Consequently the triangles HPX, WTH satisfy the conditions of w. 32 ; 
hence XHW is a straight line. 

[Mr Taylor proves this as follows : 

The projections of WH, WX on BE are \ a and \ (a + /), 
and the projections of WH, WX on 3 A are $(/-«) and |/; 
and a :{a + l)-{l-a):t, 

since aJ-/*-a t . 

Therefore WHX is a straight line.] 

(3) To prove that the pentagon BUVCW\% equiangular. 
We have BV* = BS* ■+■ SF« 

= (BN* + jVS*) + SF* 
= PN* + (NS* + SP*) 
= PN* + sPN\ 
since NS is divided in extreme and mean ratio at P [xm. 5], so that 

JVS* + J , /" = 3 /W. [xm. 4] 

Consequently By = 4PW 

=bc; 

or BV=BC. 

The As fSf, WBCam therefore equal in all respects, 
and lBUV= lBWC. 

Similarly l CVU > i.,5 WC. 

Therefore the pentagon is equiangular. (xm. 7] 

(4) To prove that the sphere which circumscribes the cube also circum- 
scribes the dodecahedron we have only to prove that, if X be the centre of 
the sphere, ZU= ZB, for example. 

Now, by xi. 38, XP produced meets the diagonal of the cube, and the 
portion of XP produced which is within the cube and the diagonal bisect 
one another. 

And ZU* = ZX , + XU* 

= IfS' + PS' 
= 3 PH\ 
as before. 

Also (cf. xm. 15) 

ZB' = ZP*+PB> 

= ZP* + PflP + NB* 

Hence ZU=ZB. 

(5) Similarly for ZV, ZW etc. 



xin. 17] PROPOSITION 17 5°' 

(6) Since FN is divided in extreme and mean ratio at R, 
NP:PR = PR: RN. 

Doubling the terms, we have 

NO:RS=RS: (NR + SO), 
so that, if NO is divided in extreme and mean ratio, the greater segment 
is equal to RS. 

Now, since the diameter of the sphere is rational, 
and (diam. of sphere)' = 3 (edge of cube)*, 

the edge of the cube (i.e. NO) is rational. 

Consequently RS is an apotome. 

SThis is proved in the spurious xm. 6 above ; Euclid assumes it, and the 
Is purporting to quote the theorem are probably interpolated, like xm. 6 
itself.] 

As a matter of fact, with Mr Taylor's notation, 
P = la + a 1 , 

and a=&— I. 

2 

Since, if r is the radius of the circumscribing sphere, >"= V3 ■ ~> 

Pappus' solution. 

Here too Pappus (m. pp. 156 — 162) finds four circular sections of the 
sphere all parallel to one another and all passing through five of the vertices 
of the dodecahedron. 

Analyst}. 

Suppose (he says) the problem solved, and let the vertices of the 
dodecahedron be A, B, C, D, E; F, G, H, K, L; M, N, O, P, Q; 
R, S, T, U, V. 

Then, as before, ED is parallel to PL, and AE to PG ; therefore the 
planes ABODE, FGHKL are parallel. 

But, since PA is parallel to BH, and BH to 00, PA is parallel to OC ; 
and they are equal ; therefore PO, AC are parallel, so that ST, ED are also 
parallel. 

Similarly RS, DC are parallel, and likewise the pairs (TV, EA), 
{UV,AB), (VR,BC). 

Therefore the planes ABODE, RSTUV are parallel ; and the circles 
ABCDE, RSTUVexe equal, since the inscribed pentagons are equal. 

Similarly the circles FGHKL, MNOPQ are equal, since the pentagons 
inscribed in them are equal. 

Now CL, OU are parallel because each is parallel to KN; 
therefore L, C, O, U are in one plane. 

And LC, CO, OU, UL are all equal, since they subtend angles of equal 
pentagons. 

Also L, C, O, U are on a plane section, i.e. a circle ; 
therefore LCOU'\% a square. 

Therefore 01? = 2LC* = zLF* 

(for LC, LF subtend angles of equal pentagons). 



5oi 



BOOK XIII 



[xm. 17 



And the angle OLF is right; for J>0, LF are equal and parallel chords 
in two equal and parallel circular sections of a sphere [Pappus' lemma, p. 138, 
30 — 26]. 

Therefore OF* = OL* + FD - }FL f . [from above] 

And OF is a diameter of the sphere ; for PO, Ft are on opposite sides 
of the centres of the circles in which they are [Pappus' lemma, pp. 136 — 8], 




Now suppose /, /, A to he the sides of an equilateral pentagon, triangle 
and hexagon in the circle FGHKL, d the diameter of the sphere. 

Then d:FL = Js>.i [from above] 

= t : k; [Eucl. xni. u] 

and it follows a/temanda (since FL -p) that 

d-.t^p-.h. 

Now let d\ p\ H be the sides of a regular decagon, pentagon and hexagon 
respectively inscribed in any one circle. 

Since, if FL be divided in extreme and mean ratio, the greater segment is 

equal to ED, [xm. S] 

FL : ED = X :£. [vi. Def. 3, xm. o] 

And FL : ED is the ratio of the sides of the regular pentagons inscribed 
in the circles FGHKL, ABCDE, and is therefore equal to the ratio of the 
sides of the equilateral triangles inscribed in the same circles. 

Therefore / : (side of A in ABCDE) = A' ;d\ 

But d:t=p:k 

therefore, ex aequali, d : (side of A in ABCDE) -p' : dl. 

Now d is given ; 
therefore the sides of the equilateral triangles inscribed in the circles ABCDE, 
FGHKL respectively are given, whence the radii of those circles are also 
given. 



xin. 17, iS] 



PROPOSITIONS i j, 1 8 



503 



Thus the two circles are given, and so accordingly are the equal and 
parallel circular sections. 

Synthesis. 

Set out two straight lines x, y such that d, x, y are in the ratio of the sides 
of a regular pentagon, hexagon and decagon respectively inscribed in one and 
the same circle. 

Find two circular sections of the sphere with radii r, /, where 

Let these be the circles FGHKL, ABODE respectively) and draw the 
equal and parallel circles on the other side of the centre, namely MNOPQ, 
RSTUV, 

In the first two circles inscribe regular pentagons with their sides respec- 
tively parallel, ED being parallel to FL. 

Draw equal and parallel chords (on the other sides of the centres) in the 
other two circles, namely ST equal and parallel to ED, and PO equal and 
parallel to FL ; and complete the regular pentagons on ST, PO inscribed in 
the circles. 

Thus all the vertices of the dodecahedron are determined. 

The proof of the correctness of the construction is clear from the analysis. 

Pappus adds that the construction shows that the circles containing five 
vertices of the dodecahedron are the same respectively as those containing 
three vertices of the icosahedron, and that the same circle circumscribes the 
triangle of the icosahedron and the pentagonal face of the dodecahedron in 
the same sphere. 



Proposition 18. 









To set out the sides of the five figures and to compare them 
with one another. 

Let AB, the diameter of the given sphere, be set out, 

and let it be cut at C so that 
AC is equal to CB, and at D 
so that AD ts double of DB \ 

let the semicircle AEB be de- 
scribed on AB, 

from C, D let CE, DEbe drawn 
at right angles to AB, 

and Jet AF, FB t EB be joined. 

Then, since AD is double 
ofZ?^, 

therefore AB is triple of BD. 

Convertendo, therefore, BA is one and a half times AD. 




S<>4 BOOK XIII [xhi. 18 

But, as BA is to AD, so is the square on BA to the 
square on AF, [v. Def. g, vi. 8] 

for the triangle AFB is equiangular with the triangle AFD ; 
therefore the square on BA is one and a half times the square 
on AF. 

But the square on the diameter of the sphere is also one 
and a half times the square on the side of the pyramid. 

[xiil i 3 ] 

And AB is the diameter of the sphere ; 
therefore AF is equal to the side of the pyramid. 

Again, since AD is double of DB, 
therefore AB is triple of BD. 

But, as AB is to BD, so is the square on AB to the square 
on BF; [vi. 8, v. Def. 9 ] 

therefore the square on AB is triple of the square on BF. 

But the square on the diameter of the sphere is also triple 
of the square on the side of the cube. [xm. 15] 

And AB is the diameter of the sphere ; 
therefore BF is the side of the cube. 

And, since AC is equal to CB, 
therefore AB is double of BC. 

But, asAB is to BC, so is the square on AB to the square 
on BE; 
therefore the square on AB is double of the square on BE. 

But the square on the diameter of the sphere is also double 
of the square on the side of the octahedron, |xm. 14] 

And AB is the diameter of the given sphere ; 
therefore BE is the side of the octahedron. 

Next, let AG be drawn from the point A at right angles 
to the straight line AB, 
let AG be made equal to AB, 
let GC be joined, 
and from H let HK be drawn perpendicular to AB. 

Then, since GA is double of AC, 
for GA is equal to AB, 
and, as GA is to AC, so is HK to KC, 
therefore HK is also double of KC. 



XHi. 1 8] PROPOSITION 1 8 505 

Therefore the square on HK is quadruple of the square 
on KC\ 

therefore the squares on HK, KC, that is, the square on HC, 
is five times the square on KC. 

But HC is equal to CB ; 
therefore the square on BC is five times the square on CK. 

And, since AB is double of CB, 
8D0, in them, AD is double of DB, 
therefore the remainder BD is double of the remainder DC. 

Therefore BC is triple of CD \ 
therefore the square on BC is nine times the square on CD. 

But the square on BC is five times the square on CK; 
therefore the square on CK is greater than the square on CD ; 
therefore CK is greater than CD. 

Let CL be made equal to CK, 
from L let LM be drawn at right angles to AB, 
and let MB be joined. 

Now, since the square on BC is five times the square 
on CK. 

and AB is double of BC, and KL double of CK, 
therefore the square on A B is five times the square on KL. 

But the square on the diameter of the sphere is also five 
times the square on the radius of the circle from which the 
icosahedron has been described. [xm. 16, Por.] 

And AB is the diameter of the sphere ; 
therefore KL is the radius of the circle from which the icosa- 
hedron has been described ; 
therefore KL is a side of the hexagon in the said circle. 

[iv. 15, Por.] 

And, since the diameter of the sphere is made up of the 
side of the hexagon and two of the sides of the decagon 
inscribed in the same circle, [xm. 16, Por.] 

and AB is the diameter of the sphere, 
while KL is a side of the hexagon, 
and AK is equal to LB, 

therefore each of the straight lines AK, LB is a side of the 
decagon inscribed in the circle from which the icosahedron 
has been described. 



So6 BOOK XIII [xm. 18 

And, since LB belongs to a decagon, and ML to a 
hexagon, , 

for ML is equal to KL, since it is a!so equal to HK, being 
the same distance from the centre, and each of the straight 
lines HK, KL is double of KC, 

therefore MB belongs to a pentagon, [xm. 10] 

But the side of the pentagon is the side of the icosa- 
hedron ; [xm. 16] 
therefore MB belongs to the icosahedron. 

7 

Now, since FB is a side of the cube, 
let it be cut in extreme and mean ratio at N, 
and let NB be the greater segment ; 
therefore NB is a side of the dodecahedron. [xm. 17, Por.] 

And, since the square on the diameter of the sphere was 
proved to be one and a half times the square on the side AF 
of the pyramid, double of the square on the side BE of the 
octahedron and triple of the side FB of the cube, 
therefore, of parts of which the square on the diameter of the 
sphere contains six, the square on the side of the pyramid 
contains four, the square on the side of the octahedron three, 
and the square on the side of the cube two. 

Therefore the square on the side of the pyramid is four- 
thirds of the square on the side of the octahedron, and double 
of the square on the side of the cube ; 

and the square on the side of the octahedron is one and a half 
times the square on the side of the cube. 

The said sides, therefore, of the three figures, I mean the 
pyramid, the octahedron and the cube, are to one another in 
rational ratios. 

But the remaining two, I mean the side of the icosa- 
hedron and the side of the dodecahedron, are not in rational 
ratios either to one another or to the aforesaid sides ; 
for they are irrational, the one being minor [xm. 16] and the. 
other an apotome [xm. 17], 

That the side MB of the icosahedron is greater than the 
side NB of the dodecahedron we can prove thus. 

For, since the triangle FDB is equiangular with the 
triangle FAB, [vi. 8] 

proportionally, as DB is to BF, so Is BF to BA. [vr. 4] 



xm. 1 8] PROPOSITION 1 8 507 

And, since the three straight lines are proportional, 
as the first is to the third, so is the square on the first to the 
square on the second ; [v. Def. 9, vi. 30, Por,] 

therefore, as DB is to BA, so is the square on DB to the 
square on BF; 

therefore, inversely, as AB is to BD, so is the square on FB 
to the square on BD. 

But AB is triple of BD ; 
therefore the square on FB is triple of the square on BD. 

But the square on AD is also quadruple of the square 
on DB, 

for AD is double of DB \ 

therefore the square on AD is greater than the square on FB; 
therefore AD is greater than FB ; 
therefore A L is by far greater than FB. 

And, when AL is cut in extreme and mean ratio, 
KL is the greater segment, 

inasmuch as LK belongs to a hexagon, and KA to a decagon; 

[xm. 9] 
and, when FB is cut in extreme and mean ratio, NB is the 
greater segment ; 
therefore KL is greater than NB. 

But KL is equal to LM \ 
therefore LM is greater than NB. 

Therefore MB, which is a side of the icosahedron, is by 
far greater than NB which is a side of the dodecahedron, 

Q. E. D 






I say next that no other figure, besides the said five figures, 
can be constructed which is contained by equilateral and equi- 
angular figures equal to one another. 

For a solid angle cannot be constructed with two triangles, 
or indeed planes. 

With three triangles the angle of the pyramid is constructed, 
with four the angle of the octahedron, and with five the angle 
of the icosahedron ; 

but a solid angle cannot be formed by six equilateral and equi- 
angular triangles placed together at one point, 



5o8 BOOK XIII [xiil 1 8 

for, the angle of the equilateral triangle being two-thirds of a 
right angle, the six will be equal to four right angles : 
which is impossible, for any solid angle is contained by angles 
less than four right angles. [xi. 31] 

For the same reason, neither can a solid angle be con- 
structed by more than six plane angles. 

By three squares the angle of the cube is contained, but 
by four it is impossible for a solid angle to be contained, 
for they will again be four right angles. 

By three equilateral and equiangular pentagons the angle 
of the dodecahedron is contained ; 

but by four such it is impossible for any solid angle to be 
contained, 

for, the angle of the equilateral pentagon being a right angle 
and a fifth, the four angles will be greater than four right 
angles : 
which is impossible. 

Neither again will a solid angle be contained by other 
polygonal figures by reason of the same absurdity. 
Therefore etc. 

Q, E. D. 

Lemma. 

But that the angle of the equilateral and equiangular 
pentagon is a right angle and a fifth we must prove thus. 

Let ABCDE be an equilateral and equiangular 
pentagon, 

let the circle ABCDE be cir- 
cumscribed about it, 
let its centre F be taken, 
and let FA, FB, FC, FD, FE 
be joined. 

Therefore they bisect the 
angles of the pentagon at A, 
B, C, D, E. 

And, since the angles at F 
are equal to four right angles 
and are equal, 




xin. 18] PROPOSITION 18 509 

therefore one of them, as the angle AFB, is one right angle 
less a fifth ; 

therefore the remaining angles FAB, ABF consist of one 
right angle and a fifth. 

But the angle FAB is equal to the angle FBC\ 
therefore the whole angle ABC of the pentagon consists of 
one right angle and a fifth. 

Q. E. D. 

We have seen in the preceding notes that, if r be the radius of the sphere 
circumscribing the five solid figures, 

(edge of tetrahedron) = | ,/6 . r, 
(edge of octahedron) - Ja.r, 
(edge of cube) = | Jx, . r, 

(edge of icosahedron) ■ - */ 10 (5 — ^5), 

(edge of dodecahedron) = - (>Jj$ — ^3). 
Euclid here exhibits the edges of all the five regular solids in one figure. 
(1) Make AD equal to zDB. 



Thus 


BA = \AD, 




and 


BA .AD = £A*;AF*i 




therefore 


BA* = \AF*. 




Thus 


AF= Vf . ar - J^6 . r = {edge of tetrahedron). 




w 


AB»:BF' = AB:BD 




Therefore 


= 3:1. 
BJ^-^Ab*, 




or 


2 2, 

BF= —7- . r m - ,/3 . r m (edge of. cube). 
v3 3 




(3) 


AB* = iBEK 




Therefore 


BE - J 3 ,r = (edge of octahedron). 





(4) Draw AG perpendicular and equal to AB. Join GC, meeting the 
semicircle in H, and draw UK perpendicular to AB. 

Then GA~*AC\ 

therefore, by similar triangles, HK= 2KC. 

Hence HK* = tKC*, 

and therefore ^KC* = UK* + KC* 

= HC 1 
= CB i . 

Again, since AB = iCB, and AD = iD£, 
by subtraction, BD= 2DC, 

or BC=sDC 



5io BOOK XIII | xiii. 18 

Therefore gUC^BC 

= $KC % . 
Hence KC> CD. 

Make CL equal to KC, draw LM at right angles to AB, and join 
AM, MB. 

Since CBP^sKC*, 

AB 1 = $KL\ 

It follows that KL (= J\,r) is the radius of, or the side of the regular 
hexagon in, the circle containing the pentagonal sections of the icosahedron. 

[xiii. 1 6] 
And, since 

zr = (side of hexagon) + i (side of decagon in same circle) 

[xiii. 16, Por.] 

AK" LB = (side of decagon in the said circle). 

But LM = HK = KL - (side of hexagon in circle). 

Therefore LM * + LB* (- BM*) = (side of pentagon in circle)* [xiii. 10] 

= (edge of icosahedron )'', 

and BM= (edge of icosahedron). 

[More shortly, HK= *KC> 

whence HK* = *KC, 

and $KC I = HC = r*. 

Also 

Thug 



AK-,-CK-r(*-^ 




BM* = HK 1 + AK* 




-!-"(;-.*)' 




= **( r ) 




= -(10-3^5), 





BM-- */io (S - Vs) = Wjf* oJ icosahedron).] 

(5) Cut />7-'(the edge of the cube) in extreme and mean ratio at N, 
Then, if BN be the greater segment, 

BN = (edge of dodecahedron) . [xi 11. 1 7 ] 

[Solving, we obtain 

BN= "^^ . BF 
a 

- Vs - 1 _* _ 

= 7<VIS-V3) 

= (fl^gv ^ dodecahedron). ] 



xm. i8] PROPOSITION 18 511 

(6) If /, 0, c are the edges of the tetrahedron, octahedron and cube 
respectively, 

If each of these equals is put equal to X, 

>*=! x, 

whence 4^* : r* : 0* : ** = 6 : 4 : 3 : 2, 

and the ratios between 2r, t, 0, c are all rational (in Euclid's sense). 

The ratios between these and the edges of the icosahedron and the 
dodecahedron are irrational. 

(7) To prove that 

(edge of icosahedron) > (edge of dodecahedron), 
ie. that MB> NB. 

By similar A s FDB, AFB, 

DB;BF=BF\BA t 
or DB:BA = DB* \ BF*. 

But iDB = BA; 

therefore BF 1 = iDB*. 

By hypothesis, AD' = 4DB* • 

therefore AD > BF, 

and, a fortiori, AL-> BF. 

Now LK is the side of a hexagon, and AK the side of a decagon in the 
same circle ; 

therefore, when AL is divided in extreme and mean ratio, KL is the greater 
segment. 

And, when BF is divided in extreme and mean ratio, BN is the greater 
segment. 

Therefore, since AL > BF, 

KL > BN, 

at LM>BN, 

And therefore, a fortiori, MB > BN. 


















APPENDIX. 

I. THE CONTENTS OF THE SO-CALLED BOOK XIV. 
BY HYPS1CLES. 

This supplement to Euclid's Rook xm. is worth reproducing for the sake 
not only of the additional theorems proved in it but of the historical notices 
contained in the preface and in one or two later passages. Where I translate 
literally from the Greek text, I shall use inverted commas; except in such 
passages I reproduce the contents in briefer form. 

1 have already quoted from the Preface (Vol. I. pp. 5-*-6), but I will 
repeat it here. 

" Basil ides of Tyre, Protarchus, when he came to Alexandria and met 
my father, spent the greater part of his sojourn with him on account of the 
bond between them due to their common interest in mathematics. And on 
one occasion, when looking into the tract written by Apollonius about the 
comparison of the dodecahedron and icosahedron inscribed in one and the 
same sphere, that is to say, on the question what ratio they bear to one 
another, they came to the conclusion that Apollonius' treatment of it in this 
book was not correct ; accordingly, as I understood from my father, they 
proceeded to amend and rewrite it. But I myself afterwards came across 
another book published by Apollonius, containing a demonstration of the 
matter in question, and I was greatly attracted by his investigation of the 
problem. Now the book published by Apollonius is accessible to all ; for it 
has a large circulation in a form which seems to have been the result of later 
careful elaboration. 

"For my part I determined to dedicate to you what 'I deem to be 
necessary by way of commentary, partly because you will be able, by reason 
of your proficiency in all mathematics and particularly in geometry, to pass an 
expert judgment upon what I am about to write, and partly because, on 
account of your intimacy with my father and your friendly feeling towards 
myself, you will lend a kindly ear to my disquisition. But it is time to have 
done with the preamble and to begin my treatise itself. 

[Prop, i.] " The perpendicular drawn /ram the centre of any circle to the 
side of the pentagon inscribed in the same circle is half the sum of the side of the 
hexagon and of the side of the decagon inscribed in the same circle." 



I. THE SO-CALLED "BOOK XIV" 



513 



Let ABC be a circle, and BC the side of the inscribed regular pentagon. 

Take D the centre of the circle, draw DE from D perpendicular to BC, 
and produce DE both ways to meet the circle in F, A. 

I say that DE is half the sum of the side of the hexagon and of the side 
of the decagon inscribed in the same circle. 

Let DC, CF be joined ; make GE equal to EF, and join GC. 

Since the circumference of the circle is five 
times the arc BFC, 

and half the circumference of the circle is the arc 
ACF, 

while the arc EC is half the arc BFC, 
therefore {arc ACF) - 5 (arc FC) 
or (arc AC)-^ 4 (arc CF). 

Hence l ADC = 4 c CDF, 

and therefore l AFC = 2 l. CDF. 

Thus 4 CGF= lAFC=x i. CDF; 
therefore [1. 31] i.CDG = i.DCG t 
so that DG-GC-CF, 

And GE=EF; 
therefore DE = EF + FC. 

Add DE to each ; 

therefore tDE = DE+ FC. 

And DF is the side of the regular hexagon, and FC the side of the regular 
decagon, inscribed in the same circle. 
Therefore etc. 




"Next it is manifest from the theorem [12] in Book xm. that the perpen- 
dicular drawn from the centre of the circle to the tide of the equilateral triangle 
[inscribed in it] is half of the radius of the circle. 

[Prop, a.] " The same circle circumscribes both the pentagon of the dodeca- 
hedron and the triangle of the icosahedron inscribed in the same sphere. 

"This is proved by Aristaeus in his work entitled Comparison of the five 
figures. But Apollonius proves in the second edition of his comparison of the 
dodecahedron with the icosahedron that, as the surface of the dodecahedron 
is to the surface of the icosahedron, so also is the dodecahedron itself to the 
icosahedron, because the perpendicular from the centre of the sphere to the 
pentagon of the dodecahedron and to the triangle of the icosahedron is the 
same. 

" But it is right that I too should prove that 

[Prop. 2] The same circle circumscribes both the pentagon of the dodecahedron 
and the triangle of the icosahedron inscribed in the same sphere- 

U" For this I need the following 

Lemma. 

" Jf an equilateral and equiangular pentagon be inscribed in a circle, the sum 
of the squares on the straight tine subtending two sides and on the side of the 
pentagon is Jive times the square on the radius." 



5«4 



APPENDIX 



Let ABC be a circle, AC the side of the pentagon, Z> the centre ; 
draw 2V perpendicular to AC and produce it to 
B,E; 

join AS, AE. 
I say that 

BA i + AC*=sDE 1 . 
For, since BE = tED, 

BE* = 4EZP. 
And £E l = BA* + AE'; 

therefore BA X -t- AE 1 + ED 1 = $E£P. 
But AC* = DE t + EA 2 ; 

[Eucl. xiii. to] 
therefore A4' + ^ C s = siXE*. 




"This being proved, it is required to prove that the same circle circum- 
scribes both the pentagon of the dodecahedron and the triangle of the 
icosahedron inscribed in the same sphere." 

Let AB be the diameter of the sphere, and let a dodecahedron and an 
icosahedron be inscribed. 

A, B 

O " 




M 



N 




Let CDEFG be one pentagon of the dodecahedron, and KLH one 
triangle of the icosahedron. 

I say that the radii of the circles circumscribing them are equal. 

Join DG ; then DG is the side of a cube inscribed in the sphere. 

[Eucl. xiii. 17] 

Take a straight line .AW such that AB I = iMN*. 

Now the square on the diameter of the sphere is five times the square on 
the radius of the circle from which the icosahedron is described. 

[xm. r6, Por.] 

Therefore MN'k equal to the radius of the circle passing through the five 
vertices of the icosahedron which form a pentagon. 

Cut MN"\n extreme and mean ratio at O, MO being the greater segment 

Therefore MO is the side of the decagon in the circle with radius MN. 

[xiii. 9 and 5, converse] 

Now $MN*=AB > = %DG*. [xm. 15] 

But 3 Z>C 1 3 C<?= sMN* 1 SM& 

(since, if DG is cut in extreme and mean ratio, the greater segment is equal 
to CG, arid, if two straight lines are cut in extreme and mean ratio, their 
segments are in the same ratio : see lemma later, op. 518 — 9). 



I. THE SO CALLED "BOOK XIV" 



sis 



And sMO 1 + $MN' = t,KD. 

[This follows from xm. to, since KL is, by the construction of xm. 16, the 
side of the regular pentagon in the circle with radius equal to MN, that is, the 
circle in which MN is the side of the inscribed hexagon and MO the side of 
the inscribed decagon.] 

Therefore %KD = 3CC + ^DG. 

But $KD = 1 5 (radius of circle about KLHf, [xm. 1 2] 

and iDG* + 3CC = 15 (radius of circle about CDEFGf. 

[Lemma above] 
Therefore the radii of the two circles are equal. 

Q. E. D, 



[Prop. 3.] " If (hire be an equilateral and equiangular pentagon and a 
circle circumscribed about it, and if a perpendicular be drawn from the centre to 
one tide, then 

30 times the rectangle contained by the side and the perpendicular is equal to 
the surface of the dodecahedron." 

Let ABCDE be the pentagon, F the centre of the circle, FG the 
perpendicular on a side CD. 
I say that 

30CD , FG - 1 2 (area of pentagon). 
Let CF, FD be joined. 
Then, since 

CD .FG=i(&CDF), 
$CD.FG = io(ACDF), 
whence 30CZ? . FG =12 (area of pentagon). 

Similarly we can prove that, 

[Prop. 4] If ABC be an equilateral triangle in a 
circle, D the centre, and DE perpendicular to BC, 
30BC. DE = (surface of kosahedron). 
For DE.BC=*(&D3C); 

therefore ^DE . BC= 6(&DBC) 
= i(&AEC), 
whence 30.D.E . BC= jo {A ABC). 

It follows that [Prop. 5] 

(surface of dodecahedron) ! (surface of icosaliedron) 

= (side of pentagon) . (its perpendicular) : (side of triangle) , (its Perp.). 





"This being clear, we have next to prove that, 
[Prop. 6] As the surface of the dodecahedron is to the surface of the kosahedron, 
so is the side of the cube to the side of the icosahedron." 



5i6 



APPENDIX 




Let ABC be the circle circumscribing the pentagon of the dodecahedron 
and the triangle of the icosahedron, and let CD 
be the side of the triangle, AC that of the 
pentagon. 

Let E be the centre, and EF, EG perpen- 
diculars to CD, AC. 

Produce EG to meet the circle in B and 
join BC. 

Set out H equal to the side of the cube in- 
scribed in the same sphere. 

I say that 

(surface of dodecahedron) : (surface of icosahedron) 

= H : CD. 

For, since the sum of EB, BC is divided at B in extreme and mean ratio, 
and BE is the greater segment, [xtlt. 9] 

and EG = ) {EB + BC), [Prop, 1] 

while EF= \BE, [see p. 513 above] 

therefore, if EG is divided in extreme and mean ratio, the greater segment is 
equal to EF [that is to say, since EB is the greater segment of EB + BC 
divided in extreme and mean ratio, \EB is the greater segment of 
\{EB + BC) similarly divided]. 

But, if H is also divided in extreme and mean ratio, the greater segment 



is equal to CA. 

Therefore 
or 

And, since 
and 
therefore 



■EGiEF, 

1 CA . EG. 



a 



H: CA-- 
FE.H-- 

CD = FE.H:FE. CD, 
F£.Sf=CA.EG, 
H.CD=CA.EG:FE.CD 

= (surface of dodecahedron) 



[xiii. 17, Por.] 



(surf, of icos.). 
[Prop. 5] 



Another proof of the same theorem. 
Preliminary. 

Let ABC he a circle and AB, AC sides of an inscribed regular pentagon. 

Join BC ; take D the centre of the circle, join AD ana produce it to 

meet the circle at E. Join BD. 

Let DF be made equal to \AD, and CH equal 
to JCff. 
I say that 

rect AF. BH= (area of pentagon). 
For, since AD ~ 2DF, 

AF= %AD. 
And, since GC= iHC x 

GC = \GH. 
Therefore FA : AD=CG: Gff, 

so that AF. GH=AD. CG 

=AD.BG 
= *{&ABD). 




I. THE SO-CALLED "BOOK XIV 



5«7 



Therefore 

5 AF. GH 
And GH= iHC; 
therefore $AF. HC 

AF.BH 



10 { A ABD) - 2 (area of pentagon). 



or 



(area of pentagon), 
(area o'f pentagon). 

Proof of theorem. 

This being clear, let the circle be set out which circumscribes the pentagon 
of the dodecahedron and the triangle of the icosahe- 
dron inscribed in the same sphere. 

Let ABC be the circle, and AB, A C two sides of 
the pentagon; join BC. 

Take £ the centre of the circle, join AE and 
produce it to F. 

Let AE=2EG, KC=$CH. 

Through G draw DM at right angles to AF 
meeting the circle at D, M '; 

DM is then the side of the inscribed equilateral 
triangle. 

Join AD, AM, which are equal to DM, 

Now, since AG . BH= (area of pentagon), 

and AG . GD=> (area of triangle), 

therefore BH : GD = (area of pentagon) : (area of triangle), 
and 1 2BH : 20GD - (surface of dod.) : (surface of icos.}. 

But i2BH=toBC, since BH=sHC, and BC=bHC; 
and 2oGD= 10DM; 
therefore (surface of dodecahedron) : (surface of icosahedron) 

= (side of cube) ; (side of icosahedron). 




" Next we have to prove that, 

[Prop. 7] If any straight line whatever be cut in extreme aud mean ratio, then, 
as is ( 1 ) tie straight line the square on whieh is equal to the sum of the squares 
on the whole line and on the greater segment to (2) the straight line the square en 
which is equal to the sunt of tlie squares on the whole and on the lesser segment, 
so is (3) the side of the cube to (4) the side of the icosahedron." 

Let AHB be the circle circumscribing both the pentagon of the dodeca- 
hedron and the triangle of the icosahedron inscribed 
in the same sphere, C the centre of the circle, and 
CB any radius divided at D in extreme and mean 
ratio, CD being the greater segment. 

CD is then the side of the decagon inscribed in 
the circle. [xm. 9 and 5, converse] 

Let E be the side of the icosahedron, F that of 
the dodecahedron, and G that of the cube, inscribed 
in the sphere. 

Then E, Fare the sides of the equilateral triangle 
and pentagon inscribed in the circle, and, if G is 
divided in extreme and mean ratio, the greater 
segment is equal to F. [xiii. 17, For.] 




E 
F 

G 



Si8 APPENDIX 

Thus E? = zBC\ [xtu. la] 

and CB* + BD» = 3CD*. [xm, 4] 

Therefore E? ■ CB 1 = {CB 1 + BIT) : CD*, 

or E* : {CB* + Bn*)=CB a : CUP 

= G*.F>. 

Therefore, alternately and inversely, 

G* : E* = f* 1 ( CB* + BIT). 

But F* = BC + CD* ; for the square on the side of the pentagon is equal 
to the sum of the squares on the sides of the hexagon and decagon inscribed 
in the same circle. [xm. 10] 

Therefore G*:E*={BC 7 + CD*) : ( CB' + BD*\ 

which is the result required. 



It has now to be proved that 
[Prop. 8] {Side of cube) : {side of icosahedron) 

= {content of dodecahedron) : {content of icosahedron). 

Since equal circles circumscribe the pentagon of the dodecahedron and 
the triangle of the kosahedron inscribed in the same sphere, 
and in a sphere equal circular sections are equally distant from the centre, 
the perpendiculars from the centre of the sphere to the faces of the two solids 
are equal; 

in other words, the pyramids with the centre as vertex and the pentagons of 
the dodecahedron and the triangles of the icosahedron respectively as bases 
are of equal height. 

Therefore the pyramids are to one another as their bases. 
Thus {12 pentagons) : {20 triangles) 

= (12 pyramids on pentagons) : (20 pyramids on triangles), 
or (surface of dodecahedron) : (surface of icosahedron) 

= (content of dad.) : (content of icos.). 
Therefore 

(content of dodecahedron) : (content of icosahedron) 

■ (side of cube) : (side of icosahedron). [Prop. 6] 

Lemma. 

Jf two straight lines be cut in extreme and mean ratio, the segments of both 
are in one and the same ratio. 

Let AB be cut in extreme and mean ratio at C, AC being the greater 
segment ; 

and let DE be cut in extreme and mean ratio at F, DF being the greater 
segment. 

I say that AB : AC= DE : DF. A C B 

Since AB.BC-AC*, 

and DE . EF= DF*, * 

AB.BC: AC* = DE.EF: DF*, 
and 4 AB ,BC:AC' = *DE . EF : DF*. 



II. THE SO-CALLED "BOOK XV" 519 

Comfanendo, 

{\AB .BC+AC*):AC* = {4DE . EF+ DF 1 ) : DF 1 , 
or (AB + BCf :AC = {DE + EFf : DF 1 ; [li. 8] 

therefore {AB + BC) : AC = {DE + EF) : DF, 

Compotundo, 

{AB + BC + AC):AC = {DE + EF + DF) : DF, 
or 2AB ': AC= xDE : DF; 

that is, AB : AC= DE : DF. 



Summary of results. 

If AB be arty straight line divided at C in extreme and mean ratio, AC 
being the greater segment, and if we have a cube, a dodecahedron and an 
icosahedron inscribed in one and the same sphere, then : 

(1) (side of cube) : (side of icosahedron) = J{AB 1 -i-AC 1 ) : ,J {AB* + BC); 

(2) {surface of dod.) : {surface of icos.) 

= {side of cube) : {side of icosahedron) ; 

(3) {content of dod.) : (content of icos.) 

= {surface of dod.) : (surface of icos.) ; 
and (4) (content of dodecahedron) : (content of icos.) 

= J{AB> + AC*) ; ^{AB' + BC). 



II. NOTE ON THE SO-CALLED "BOOK XV." 

The second of the two Books added to the genuine thirteen is also 
supplementary to the discussion of the regular solids, but is much inferior 
to the first, " Book XIV." Its contents are of less interest and the exposition 
leaves much to be desired, being in some places obscure and in others 
actually inaccurate. It consists of three portions unequal in length. The 
first (Hetberg, Vol. v, pp. 40 — 48) shows how to inscribe certain of the 
regular solids in certain others, (a) a tetrahedron (" pyramid ") in a cube, 
(A) an octahedron in a tetrahedron {"pyramid"), (^) an octahedron in a cube, 
(J) a cube in an octahedron and {<) a dodecahedron in an icosahedron. 
The second portion (pp. 48 — 50) explains how to calculate the number of 
edges and the number of solid angles in the five solids respectively. The 
third {pp. 50 — 66) shows how to determine the angle of inclination between 
faces meeting in an edge of any one of the solids. The method is to con- 
struct an isosceles triangle with vertical angle equal to the said angle of 
inclination ; from the middle point of any edge two perpendiculars are drawn 
to it, one in each of the two faces intersecting in that edge ; these perpen- 
diculars (forming an angle which is the inclination of the two faces to one 
another) are used to determine the two equal sides of an isosceles triangle, 
and the base of the triangle is easily found from the known properties of the 
particular solid. The rules for drawing the respective isosceles triangles are 
first given all together in general terms (pp. 50—5*); and the special interest 
of the passage consists in the fact that the rules are attributed to " Isidorus 



S'o 



APPENDIX 



our great teacher." This Isidorus is no doubt Isidorus of Miletus, the 
architect of the Church of St Sophia at Constantinople (about 532 a.u.), 
whose pupil Eutocius also' was; he is often referred to by Eutocius {Comtn, 
on Arckimeiits) as o MtAijVtos ftnvamfSs 'l<riStopoi if/itTipos &iMaxa\tK. Thus 
the third portion of the Book at all events was written by a pupil of Isidorus 
in the sixth century. Kluge (Dt Eudidis eltmetUsrum libris qui ftruntur XIV 
tt XV, Leipzig, 1 891) has closely examined the language and style of the 
three portions and conjectures that they may be the work of different authors; 
the first portion may, he thinks, date from the end of the third century (the 
time of Pappus), and the second portion too may be older than the third. 
Hultsch however (art. " Eukleides " in Pauly-Wissowa's Real-Eneyclopddie dtr 
dassischtn Altertumrwissenuhaft, 1907) does not think his arguments con- 
vincing. 

It may be worth while to set out the particulars of Isidorus' rules for 
constructing isosceles triangles with vertical angles equal respectively to 
the angles of inclination between faces meeting in an edge of the several 
regular solids. A certain base is taken, and then with its extremities as 
centres and a certain other straight line as radius two circles are drawn ; 
their point of intersection determines the vertex of the particular isosceles 
triangle. In the case of the cube the triangle is of course right-angled ; in 
the other cases the bases and the equal sides are as shown below. 



For the tetrahedron 
For the octahedron 
For the icosahedron 

Fur the dodecahedron 



Bast of isgsttlts irianglt 
the side of a triangular face 



the diagonal of the square 
on one side of a triangular 
face 

the chord joining two non- 
consccutive angular points 
of the regular pentagon on 
an edge (the "pentagon of 
the icosahedron >N ) 

the chord joining two non- 
consecutive angular points 
of a pentagonal face [BC 
in the figure of Eocl. XJIi. 
■7l 



Eqitat sides of 
isttsteles trinnglc 

the perpendicular from the 
vertex of a triangular face 
to its base 

ditto 



ditto 



the perpendicular from the 
middle point of the chord 
joining two non-consecu- 
tive angular points of a 
face to the parallel side of 
that face [NX in the figure 
of Eucl. xiti. 17] 















GENERAL INDEX OF GREEK WORDS 
AND FORMS. 






[The references are to volumes and pages.] 



<kywnor t angle-less (figure) t* 187 

Aivfarov : f? c it t6 &.&. airuytiryf), rj StA. roi' &5. 

fat^it, jj fiirj i J. &yov<ra drd&eifif L 136 
(Ui3o<«5^j p bar^-hke 1. 18S 
Axj»t, exlreme (of nuriiliL-[>. ill a series) IE. 
m <i$, 367 : (fit} axpw *aJ ^ Vok X ri v * 
T£Tfifyrdat t "to be cut in extreme and 
mean ratio" it. 189 
iXo^ot, having no ratio, irrational JL 117-8: 
a relative term, resting on assumption or 
convention (Pythagoreans) lit. f, 11: use 
of term restricted in Euclid 111* 13 
avfiXt'ta j-fuWa), obtuse (angle) [. t8i 
&fifi\trr<itvun, obtuse- angled 1. J 87 
djuepifr, indivisible t. 41, 268 
A.fi.'iti.KOiXct (of curvilineal angles) l. 178 
Afi$txvfmi (of curvilineal angles) J. 178 
ar*ypdifKu> Att6, to describe on, contrasted 
with to construct {owrf}oa00ai) I. 348 ; 
peculiar use of active participle, a! fcra 
Ttrp&ytoHHt &vaypd4x}\iv at = straight lines on 
which equal squares arc described 1 J t- 13 
-:ii-fi\-:'>-'a, proportion : definitions of, inter- 
polated H, no 
dr dAoTc*' = <i*a A67W, proportional or in pro- 
portion; used as indeclinable adj. and as 
mtv. n» 119, 163: "i-.r^. ti ;■■ -il N ■>> :.■ , mean 
proportional (of straight JineJ if. 119, 
similarly ftiuot dwdAflTO* of numbers ](. 
305, 363 etc : TpLrt} (rph-or) dvd\oyCt>, 
third proportional 11, 314, 407-8: rerdprij 
{Tfaaprvt} ay&Xoyov, fourth proportional 
XL fiji, 409: ii:r;$ Avdhoyw^ in continued 
proportion ][. 346 
draXytyurpj (r^rcr), Treasury of Analysis, 

1. 8, 10, ii t 138 
drta-aXc (\^70f} r inverse (ratio), inversely u. 

134 
&*ajrTpt$am, convertendo t in proportions ri. 

133: analogous use otherwise than in 

proportions m. 164 
dwrps^Tr \Ayov t "conversion " of a ratio 11. 

"35 

AratirpofaK&t (species of locus) 1. 330 



Apicdxtt dvtfdKit (trot, unequal by unequal 

by equal (of solid numbers) — scalene, 

irtfnjwtffKoSi tF^n}KtffKot or fiufxlaKos 11+ 390 
dPOjUHtyi<inH, non-uniform I. 40, 161-3 
dvOjULoJuf Tfra^^i'afi' twk X^yux (of perturbed 

proportion) in Archimedes 11+ 136 
avramlptmi, ^ ai>r^, definition of same ratio 

in Aristotle {dvdif^aiptcrtf Alexander) U, 

130: terms explained II. ill 
d*TnrtTov86Ta. axn^ara, reciprocal ( m recipro* 

cally related) figures, interpolated dcf. of, 

II. 189 
dyTiffTparfrff, conversion 1. 156^7 ; /coding 

variety, ij rpoiryfltyifrij Or tj Kupius, ifa'd* 
AyuwapxTQi, non-existent t» 119 
ci^u.'j', axis lti. 369 
a.&paFTOi, indeterminate : (of lines or curves) 

I. t6o: (of problems) 1, 139 
dro^wyif, reduction 1+ 135:' c^s Ti ddtfrarnp 

1. 130 
tlrttpo?, infinite: ^ ttr* Air. tV^aWsmr^Tj of 

line or curve extending without limit and 

not H forming a figure" I. 160- 1 : fr' dr. or 

elr dr. adverbial lr rgo + ir dr+ Aiatptfirfai 

I. 368 : Aristotle on to Avupov r. 333-4 
dirXanff, breadth less : in definition of a line, 

fJjxiK drXar^T, breadthless Length I, 158: 

(of prime numbers) IT. 383 
drXafa, simple: (oflines or curves) i. i6t-3: 

(of surfaces) I. 170 
drofoifrf, Z 11 ^ (one of necessary divisions of 

a proposition) 1. 139, 130 
d jto tanner rarurfc* recurrent {—spherical}, of 

numbers tr. 391 
ci.woTOfj.ri, apofomc, a compound irrational , 

difference of two terms m. 7 : defined m. 

158-0: ^ffijr droro^ rptinj (Sew^po), 

first (second) apototne of a medial (straight 

line) in. 7, defined in. 159-60 
drrfffffa*, to mgfi, occasionally 10 touch 

(instead of i^iwrto&ai) t, 37, II. 3: also 

= to pass through) to lit on It. 79 
dpefyuta, number definitions of, u, 380 
dpprjTM, inexpressible! irrational: of X<5tot 



5*2 GENERAL INDEX OF GREEK WORDS AND FORMS 



|» 137 : ftppijTor Jidjurpaf itjt rrpvate, 
11 irrational diameter of 5 " (Plato) = */fio, 
1, 399, in. 12, 515 
ApTiaxis dpT-ioSiVa^tjir (Nicornachus) It. 28 1 
.:!/■ j- l &Vir ajjrtoj, even- times even n. 281-1 
dpridViv Trepitrffifj even-times add EL 2S2-4 
dpwoirlpcrroF even-odd (Nicornachus etc*) U. 

Airier {ilj.^V'H' r - V|J[l (number) II- 281 

dtfWU/Jar.,, :■!:■. -I'ii| :■'!".!..■ I. !.'..,* 

&<rvftfn*Tpos, incommensurable: d +J utj*** (^.hvor) 
incommensurable in length (only), 5i<vafj.it 
"in square" III, 1 [ 

dff^iTTtirroj, not -meeting, non-secant t asym- 
ptotic I. 40, ifH. 303: (of parallel planer) 
III. 165 

Mr0*T«t t incomposite: (of lines) 1. 160, 161 : 
(of surfaces) I. 170: ( prime and) ^com- 
posite (of numbers) It* 384 

aVa*T«, unordered: (of problems) 1. 128; 
(of irrationals) i, n;, lit 10 

dro^i 'vpaftuaJ, "indivisible liues" 1. 268 

£dtf« t depth 1. 158-9 

r\;\ -.-.', base 1. 2i> : -fJ 

fltfiyxiimti to stand (of angle standing on 

circumference) tt, 4 
ptafitffKiH, altar-shaped (of "scalene" solid 

numbers} IX, 190 

7f7wArw (in constructions), "let it be {have 

been) made" EL 148 
>f7wiT A> ffij ri jg&oyMr, " what was 

enjoined will have been done" it. Bo-, ?6t 
7f>f>a0*w 1 *■ let it be (lit. have been) drawn " 

l. 141 
'yfvfyiei'or* 6 ij- afrrwp, " their product H u. 

3161 326 etc* : 6 4k rou rot "f&tip&m 

~ ,4 the square of the one" It- 317 
fME^tNT] gnomon a,v*: Dcmocritus i-epf &ux- 

<popys yvufioyos fortbjPfl or yiavl^ ?) ^f fffpl 

Vnr'^LUT ftfoXoV Kill <?tpa.iyT(% If. 40; (of 

numbers) It* 289 

' i ;-<jt\"ii, line -{or curve) #*v. 

ypufituxbt, linear (of numbers in one dimen- 
sion) li. 187: {of prime numbers) !!♦ 285: 
7pauuut£t, graphically I, 400 

7pctyra'9w, "to be^r«wrf" (Aristotle) ti. no 

Srfo^frof, given i different senses t, 131-3 : 
Kuclid's StSof^va or Data q.v* 

B&jfpQMt, illustrations* of Stoics I- 329 

ffiT ffij, "thus it is required hp (or "is neces- 
sary"), introducing foaparpbi t. 293 

fctfrcpoij secondary (of numbers); in Nico- 
rnachus and lamblichus a subdivision of 
odd 11, 186, 3S7 

8*X&f*fvov, "admitting" (of segment of circle 
admitting or containing an angle) U. 5 

ftur-HH^xa = proposition (Aristotle) ]■ 252 

&iaipfiix#cu (used of "separation * T of ratios) : 
d'uupf t?frra, sepuranda^ opp. to a t^yjcrl/i*^ a, 
contponetsdo II. 168 

dtaipeaw, point of division (Aristotle) I. 165, 
170, 171 : method of division {exhaustion) 



I. 185 : Euclid's Ttpl duupfreur, 0« rfrfri- 

stons [of jigures) f. ^, y, is, ^^ no; &of- 

pfffir XtVyou, separation* literally dnnswn t 

of ratio ti. r 35 
iuiitfTpQr,. diameters of a circle, parallelogram 

etc. r. 185, 315; of sphere hi, 170 
fffurdtfftf, almost = ^dimensions " (. i$f r 

1 j8, i"i. 363 : Aristotle speaks of six A il 203 
fcaararAp, extended* iff iw one tvay t 4*1 Stfo 

two ways, iwl rpt* three ways (of lines, 

surfaces and solids respectively) 1* 158, 

(70, in. 263 
Suwrypci, distance I- i66> 16 j, 107 : (of radius 

of circle) 1* 199 *. (of an angle) = divergence 

r, 176-7 
dufttfYttJvv [&vtL\oyta), disjoined = discrete 

{proportion) ir. 393 
&it\6vTt, separando t literally dh/idendo (of 

proportions) U, [35 
$tt£o6tK&t (of a class of loci) [. 330 
Sijlptjii £rr$ { foaXayla) , discrete [ proportion)* i .e. 

in :-.nir terms t as distinct from continuous 

{awiX$f* rvryijipdni} in three terms II. I}|, 

ffi7Jx^w T "let it be drawn through" { = pro- 
duced) or "across" 1, *So* 11. 7 

5t" tvc<\ . ex aequati (of ratios) ]t + 136: &' 

Mf ■■11. ^»r r . r cf .-:zy:.< : T ;■;■ ■•£;■;].>, !.. y;'a ; ' ' CX tiSquaii 

in perturbed proportion" TI. 136 
3urAXDv l Fwf» fwice-truntated (of pyramidal 

numbers} ll. 391 
<5[0/>i<r/ri)f = |i) particular statement or *lcfini- 
tiuiu one of the formal divisions of a pro- 
position [. 119: (ij statement of condition 
of possibility h i?8> 12^, j jo, 131, 234, 

^ 43. *93 

^■■Xiirioi \-Jyes , detiA/c ratio: o'tirXatrJui' Xot«, 

duplicate ratio, contrasted with, [I. 133 
[Wa^tf, power: —actual value of a sub- 
multiple in units (Nicomachus) it, t%t\ 
= side of number not a complete square 
[<■■■. root or surd) in Plato II* i&8, 190^ 
m. i T it 3: —square in Plata it. 294-5 
i>v-i--a.TfMi, "10 be side of square equal to" 
111. 13 : a! fiuyafi.tva.1 a^i, sides of squares 
equal to them [it. (3: jj Bl 1 tjji A pttfoir 
tifanrtit rjj A%, "the square on BC is 
greater than ihe square on A by the square 
on DF* literally " BC is m power greater 
\Y\nn A ty DF ,f ail 43 

tl&bt, figure ll* 234: -furjti n. 354 
ffja^uTi) (ij.iA. di-£a- r j. Introduction ta tVarmonyt 

by Cleonides 1. 17 
ixaffTQi, each 1 curious use of, 1 1. 79 
tKarfjta. inn.rip^ t meaning respectively 1- 248, 

tKflfjihfyrtiuen,*, use of t 1, 244 

'.Vclj-.t- Im^lMiI i, 400 

fxflwii, setting-out, one of formal divisions 

of proposition r. 129: may sometimes be 

omitted I. 130 
^«r6f T sari to (of an exterior angle in sense 

of re-entrant) t. v6y* 1} #Vr« >aw/a, the 

exterior angle I. 280 



GENERAL INDEX OF GREEK WORDS AND FORMS 523 



tXaacrue, mirror (irrational) straight line ill, 

7 etc 
ikmtt&fa, spiral-shaped l. 159 
tXkt it*?*, deftet [in Application of areas} 11.161 
fWiiTtiy, ''fall short" (in application of 

areas) ij. 162 
fkXtitfftt, fatting- short (in application of areas) 

I" 3°> 343^5p 363-+ 
AA*f£i lywDUpil, a deficient (indetermi- 
nate) problem j. 119 
IftiriTTt&ifa// in { — be interpolated) It* 358 
iwabM$ t alternately or (adjectivally) alternate 

1. 308: iwaXKai Myotj alternate ratio, 

alttmando it. 134 
fro rXrtw, "several ones'* (def. of number) 

it. 3B0 
4t*pp6£rw t io fit in (active) Book IV. Def* 7 

and Prop. 1, it, 79, So, 8t 
(rraa, notion t use of, t. 111 
fpffTUrtr, objection I. 135 
frrir, within : (of internal contact of circles) 

It. 13: ptord tA JitIj or ^ irrot (>uWa), of 

an interior angle [. 263, 180: 4 rf*-r£r aoX 

AwerarTiow yurLa., the interior and opposite 

angle r. 380 
4f*jt avdXpyor, in continued proportion (of 

terms in geometrical progression} It- 346 
^■■efetfxtfttffw {^rifriryvi^u, join) 1. 341 
/T^p4£M )i6yot,fu/>erparticufaris ro/w — ratio 

(* + i):it, ii. 39$ 
/rfrtAor, j§/jw in Euclid, used for surface 

also in Plato nnd Aristotle I. 169, m. 263 

Mon ft i {dfxipAs), plane (number) ft 187-8 

irirpoffB&r, irtwpoo-8tv thru, to stand in 

front of (hiding from view), in Plato's 

definitions of straight line and plane l« 

165, 166 
trifuwo, surface 1 in Euclid L 169 ; in 

Aristotle III. 163 
eir&ttfra f consequents (—"following" terms) 

in a proportion R. 134, 138 
irtpoti^$CTjf t oblong. ertp6fiTjKti t oblong {figure) 

I* 151, r88 ; (of numbers) in Plato - wpo 

fiqxqt, which however is distinguished from 

frfpc^t7Jvnp by Nicom&chus etc- ll. 289-90, 

293 
tfflv* rA, the straight j. 159: t&faui. (7/w^tf), 

straight line 1. 165-9 
ci>$vypafituic6t t rectilinear (term for prime 

numbers) [I. 185 
tvQ<ryfmp.fi.oi t recti lineal 1. 187: neuter as 

substantive I. 346 
itih-prrpticai, euthy metric (of prime numbers) 

& 2S5 
itfnvTtiTtfat. to touch [* 57 
{<papy.u'itiv. to coincide, {$*ptt&ft<j8ai, to be 

applied to \, |6&, 214-5, 249 
tftxrixto (of a class of loci) I. 330 
i<p*i%t 7 *' in order " i. t8j 1 of adjacent angles 

J. [81, 278 

iryo&fMta t antecedents {"leading" terms) in 

a proportion it. 134 
ftrep, 1 ban: construction after &iw\aei-tiii> etc. 

K. 133 



tffijp?j,ua b theorem, £.z>, 
0^7*6* (shield) = ellipse I. 165 

l&iatt.-JixTjtj of square number (lamblichns) U» 

193 
i'TTfli' jt^Sij, kippopede (horse- fetter), name 

for a certain curve I, 162-3, x 7" 
iaaxts tcr Axis foot, equal multiplied by equal 

and again by equal (of a cube number) 11, 

390, 191 
iffdms Kror, equal multiplied by equal (of a 

square number) II. 39r 
la&ttt too* IkaTTQvdKii {p£t£a*&Mi), species of 

solid numbers, =Tt\u'&it [Hath or orijXH} 

tt 291 
UauiTftuiv <ivr,^riTi.ov. reply On isometric 

figures (Zenodorus) Z. 16, 27, 333 

/c&Strot [fv&da ■ypaftfuft) t perpendicular L 
181-2, 371: "plane" and ** solid" per- 
pendicular T. 373 

KadijyTrfiit 1. 10 

tcaXtic&w, ** let it be called," indicating 
originality of a definition II. 129 

xap,rv\os, curved (of Lines) ]. 159 

xaratitTiuh', measure II. u 5: without re- 
mainder, *' completely Jt {T\flp«J?rcrt) ll, 280 

naTotfKevA^v-, construct : rue it&rum rara- 
vxiuacQisfTutv*. "with the same construc- 
tion" n. 11 

Kareufm i -J. construction t or machinery, one 
of the divisions of a proposition I. 129 : 
sometimes unnecessary 1* 130 

icarartyiq Ka.r6yo%, Settio canenis attributed to 
Euclid [. 17, 11. 295 

vtltr&to, "let it be made" i. 169 

xf*:atiu.*vr ! , bent (of lines) I. [59, [76" 

KfrTpar, centre I- 183, 184, loo: of sphere HI. 

270: ^1 iljE Tf»i xirTpOt p = radiUS l. TOO. IT. 2 
xepiLTQ€t&itt [yvt>la) t horn-like (angle) L 177, 

178, 1S3, H. 4, 39, 40 
irXtt*-, to break off \defit(t t or inflect: Kfk\d<F&tu, 

def. of t alluded to by Aristotle t- 118, 150, 

t?6, [78, II. 47 : KtJcXanrfrfvii ypap-fi^, 

defined by Heron I. 150, 159: jratXitftfw 

Jty TdW M. 47 
A-Xdffit t breaking (of lines) 1, 176 
r\^(t, inclination: (of line to Line) 1. 176: 

{of straight Line to plane or of plane to 

plane) [II. 263-4: i>tiatw jcejcAfr&u, to be 

similarly inclined III. 165 
imXo-nLviov, hdifo'ii'rittgicd (figure), in Zeno- 

dorus t. 37, 188 
KQtvoii itrottH, Common Notions (^axioms) 

I. ui j j; called also ret kqu>&, Kauai &bj-at 

(Aristotle) I. no, 221 
xoirii rpoffxttcffu, atfwnaOu, "let there be 

added to, subtracted from, each" I. 276 
*.ot j- ij Ti\L<.r ls common section (of planes) ill, 

263 
KoXuvpotj truncated (of pyramidal number 

minus vertex) tl. 291 
xapvaty, vertex: xara. xopvif^P, vertical (angles) 

1. 278 
xpixot. ring (Heron) 1. 163 



534 CtENERAL INDEX OF GREEK WORDS AND FORMS 



fcwc\«if, eydic t a particular species of square 

number ll- 391 
KvXir&pos, cylinder itl. 3.71 
xu>vQt t cone HI. 370 

XW«h lemma (■= something assumed, Ao>*- 
pard/aw) E 133-4 

>*>«, ratio: meaning H. 117: definition of, 
if. 116-9: original meaning (of something 
expressed) accounts for use of AXpyoF, 
having n$ raliOj irrational II. 117 

Xotvfa, remaining: *«t^ j& A A \tnrjJT$ BH 
Bff-sj lm& j. 345 

^fii'^i', major (irrational straight line) 111. j, 
S7-8 etc, 

fittiorOffBtUf to be isolated \ of /iwdt. Unit 
(Theon of Smyrna) II* 279 

fdpos, part: two meanings II. 1*5: generally 
= submultip]ell*38o: p$p% parts ( = proper 
fraction) 11. 115, 180: pipy { = direction) 
]. 190, 308, 333: (=side) t, 171 

ttfa), mean proportional (straight line or 
number) II. mo, ■!■<>-. 363 etc* 

uJ^as, "medial (of a certain irrational 
straight line or area) lit. 49, ;o: 17 t\ Svu 
^tfwv TTpifrrit {Stvripa}, "the first (second) 
himedial (straight line)" 111. ;, 84-6: 
ftdfftji dworofiif wpiirnf [tkvrfpa.), "first 
i second) apotome of a medial (straight 
line) 1 ' III. 7, 159-63: fartr irai fUro* 
c\vn,u 1 i-"ij t "side of (square equal to) the 
sum of a rational and a medial area" 
HI* 7p 88—9: 3i/o pfoti Hvrafilmj, t4 side of 
the sum of two medial areas" Hi. 7, 
89-90: i} ptrft pjroO f>cfraLi) utfj-Ji- ri oW 
iroioutra, "side of {square equaj to) the 
difference between a medial and a rational 
(medial) area ill. 7 t 164-7 

fttrttiipoi, elevated (above a plane) in, iji 

M'7*fy>* "suppose it is not*' II. 7 

WW, length 1. 158-9: in Plato = side of 
complete square or length commensurable 
with unit of length 11, 388, ill. 3: more 
generally, of number in one dimension 
II* 287-8 

fntvoti&/j; t tune-like (of angle) I* i6 t -201 ; rh 
t±T}»Qti$h (cr^^ta), Lune I. 187 

)■■■;■.:■.■■:. " mixed" (of lines or curves) l. i6i 1 
161: (of surfaces) t, 170 

paras, unit, monad : supposed etymological 
connexion with ,tii;-ji r solitary, p.w$ t rest 
11.379: pwai jr l :..:,,7.' l .a.^ni' t-a ('(iru', definition 
ofbpoinf 1. 155 

^ui^crTpa^fff Ad;, " single-turn spiral " 1* 1 1 2 - 
3 «., 164-5 : m Pappus — cylindrical helix 
i- 1 65 

itrciri, indi notions t a class of problems 
1. 150-1 : rtfaut, lo vtrge 1. 118, 150 

{wr/weijijii, scraper-like (of angle) 1* 178 

6&oti6$t, "*>f the same form" 1. 350 



6^ioio^^ T uniform (of lines or curves) E, 40, 
161-2 

JS>i«« P similar; (of rectilineal figure*) il. 186: 
{of angle*) = equal (Thales, Aristotle) l* 
353 : (of segments of circles] II. 5; (of 
plane and solid numbers) l, 357, \\. 393 

afioi^TTjs \6yaift "similarity of ratios " (inter- 
polated def. of proportion) 11, 119 

Apiktryot, homologoui* corresponding II- 134: 
exceptionally "in the same ratio wiih " 
It- 138 

6vbiAa T name or term, in such expressions as 
i} tn fifo AvotidTw, ibe binomial {straight 
line) tit. 7 etc. 

4£ti& i-fuvla}, acute (angle) f. 181 

:■.;.-> iL<rw; t acute-angled I* 187 

Svtp l&ti fcii;a.t (or rouprai) g. k.e>. (or F.) I* 57 

6p8<ry&t>t<rt, right-angled : as used of quadri- 
lateral s = rectangular t . 1 88-9 

optirfi&f, definition I* 143 

Spot, definition I* 143 : original meaning of, 
l. 143: —boundary, limit I. 1&3 : =term 
in a proportion jl 131 

o^/ijj visual ray 1* 166 

i-acT-p itfTtLkattpariittrat. "taken together in 
any manner t. 318a 

mpafidWe^ to apply tan area) : T&pa$6Xk*\r 
d .t 6 used, exceptionally, instead of -wapa- 
fidWttv -wapii or ayayp&ifrtvr &r6 II- 363 

wapafloXi} run? xtipltuvi application of areas 
l - 3^> 343~5 : contrasted with inrtpfiokii 
Uxceedtng\ and fAV^i {Jatimg^hort) 1. 
343: mpafioK^ contrasted with cvarafit 
{construitian) I* 343 : application of terms 
to conies by Apollonius i. 344-f 

irapd^ofar r^wtn, a, The Treasury of Para- 
doxes J. 339 

xapaXMTTw* "fall beside," H sideways" or 
"awry" % 267, it. 54 

■wapaXk-tiKtitiridot (ad j . ) , parallel e pi pedal = 
*'wilh parallel planes or faces": areptor 
TrapaWifXttriiredaif = "■ parillelepi pedal 
solid* 1 ' not "solid parallelepiped" ML 336 

wapa.Wyf\6ypafi.fi.oj, parallelogram mic {= pa- 
rallel-lined) : TrafmW-rjX/jyfiatitxoy K^pio* 

41 parallelogramm ic area," shortened to 
Tapa\\Tj\6ypapyLQy r parallelogram I. 335 

wapa-wXitpWfUi^ eomplement (of a parallelo- 
gram) y.v. 

TrcvrdypafAftor 11. QX^ 

■jripaiirmma jrofl-Arijr, "limiting quantity" 
(Thymaridas* definition of unit) 11* 379 

w/pai, extremity 1. 165, 181 : wJ pai v u-.-xXtto* 
(E } osidonius' definition of Jtgttre) 1* 183 

TtfXfXQjUwTj (of angle}, w<pit\6fitrQr (of reCt- 
angle)^ contained h 370) rA oh wept*x&- 
".(S--UI-, iit'ice the rectangle contained 1*380: 
(of figure) contained or bounded I* 183, 
183, 184, 186* :S 7 

ir<^njcrdKiT dprtott odd-times even 11, 183-4 

irzpureaKtt vf^aaa^ cdd-tlmes odd UU 184 

TriptfadyTMt, odd-even (Nicomachus etc.) 
ir. 983 

wtptets6t t odd (number) II, 381 



GENERAL INDEX OF GREEK WORDS AND FORMS 525 



mpi&fpeiai circumference (includes are) 1. 184 

■rtixiptpTj's, circular 1- 159 

vtpi$€p6ypij.fLpoi, contained by a circum- 
ference of a circle or by arcs of circles 
i. (82, 184 

xijXiVfli, how great I refers to continuous 
(geometrical) magnitude as rots fa to discrete 
(multitude) II, 116-7 

mjAucAnn, used in v. Def. 3 and Vt. Def. 5 : 
=siu (not quantuplkity as it is translated 
by De Morgan) 11. 116-7, 189 ; P : sup- 
posed multiplication of xiyXtjrAfjjrtfj (VI. 
Def. 5) 11. 131: distinction between 
ir7j\«dTijt and iUytBoi H- 117 

rXdrof, breadth I. 158-9: (of numbers) tl. 288 

t\- ,u iL^.i- 1 -r^.\:^i. : i, m (problem) in excess " 
]. 139 

»\ft'pi, aide : (of factors of " plane " and 
" solid n numbers) ll* 1B8 

rKijSoi Ctptfffiiroy or Ttrtpairu&Q*, defined 
or finite multitude (definition of number) 

I L. l&O : 4x fi.QV<i5wv o\ryKilt±tvuy irXifBot 

(Euclid's def.) n. 180 

*BX\(m:\a<r\<ii'tti', multiply ; defined II. 287 

*e\\a.T\aQi&0frfit t multiplication: tcati* 6>oi- 

woO* TroX'\an\a.sta/jfi6f, " l (arising) from any 

multiple whatever" U* 1 20 

To\Xair\djior, multiple : ledums -woWaTXiLtTta, 

equimultiples ir> no etc. 
vtoot, a mathematical instrument 1. 370 
vpXurtaupop, multilateral, many-sided figure 
1* 187 : excludes rerpdrtaiffer, quadri- 
lateral II, 239 
Topto-offffoi, to "find" or H furnish" 1. 125, 

IT. 2 4 8 
r6ptfffta t poiism q t v* 
rardjui rptr&Ktt twoJ, **so many times so 

many times so many" (of solid numbers, 

in Aristotle) It* 286, 290 
irordttf rwaf, " so many times so many " (of 

plane numbers, in Aristotle) 11. 286 
ww6r, quantity y in Aristotle It* 115: refers 

to multitude as T-r}\\.Km to magnitude II. 

1 16-7 
rfiUffta, prism in. 268 
rpd^Xujto, problem £.f. 
Tp«77fH/^Lfi'0T p iutdiitg: (of conversion) = 

complete E. 256-7* rpQirjtQ&pevor (8c\i/pT}iia). 

leading (theorem^ contrasted with converse 

■pojufiriff, ffWew^ (of numbers) : in Plato 
=^Tfj»^)JifUt t tut distinguished from it by 
Nicomachus etc. n. 289-90, 293 

rpfa, in geometry, various meanings of, 1. 277 

wpovvvypdfai, to draw on tox (of a circle) to 
complete, when segment is given n. 56 

Tpoffapti&fovaa {tvBria) = '+ annex," the straight 
line which, when added to a compound ir- 
rational straight line formed by subtraction, 
makes up the greater "term,'' i.e. the 
negative "term ill* 150 

rpovtvpeir, to find in addition (of finding 
third and fourth proportionals) n» 214 

rpbravtt, enunciation T + 129^30 

rpvrdPUt to propound f« 12B 



wpoTiQ4v<Ui to propose : tj TporcBcura rttfrict, 
any assigned straight line in* ir 

TpwT<n Tpot iXX^Xour, (numbers) prime to 
one another II. 285-6 

rpwroT, primed two senses of, !♦ 146: 11*284-5 

xriffir, cast 1. 1.J4 

wvpafiii, pyramid tU* 268 

fnjrhj rational (literally "expressible") I. 
137, II. 117, 111. 1 : a relative term, un- 
like dati/AfUTpas (incommensurable) which 
is a natural kind (Pythagoreans) HI- 1 : 
jiT-tjT-Tj frd/icrpo; 7H7T Ft^jrdSo?, " rational 
diameter of 5 " ( = 7, as approximation to 
V50J t. 399, Rf, t2, £2£: ^rop Kat foiffav 
Swafrfr/f ( = side of square equal to sum of 
a rational and a medial area) etc itl. 7 

:.r ?;■:.■■•; '-!;■_ point !• [ £5 — 6 

'TTL-t.v??. a mathematical instrument I* 371 

vreptfa, solid Itl- 262-3: of solid numbers 
II* 290-1 : vrtpta yuryia., solid angle ML 
267-8 : fjuoia arcped fxV flf *i similar 
solid figures HI. 265-7 

irrvy^ point l. 156 

(TTo<,x*f L 0i', element J. 114-6 

ffrpoyyi/Voi', tA, the round (circular), in Plato 

I, 1 59. 184 
VTpoyyu\&Ti}t, roundness 1. 182 
trutipcTpoty commensurable : M** 1 * in length, 

S\ vii;.n ! .arj^o:', in square only 111. 1 1 
■rfuiTrfpao-ua. conclusion (of a proposition) 

[. 129, (30 
riyemf, convergence L 282 
ffirex^f, continuous : aw*xk* AraKoyta, 

"continuous proportion " (in three terms) 

II. 131 

. t 1 ;■ , i : , .■ . a.- :-i dVaXcyfo P connected ( i + e- continuous) 
proportion II. 131, 293: wmf^ftdvot of 
tonipound ratio in Archimedes 1 1, 133 

*w4Mrvi| c&mponettdo n. 134-5 

o-L^dWir XAyflu, '^composition of a ratio, 11 
distinct from compounding of ratios TL 

ffirp-^ro^ composite: (of lines or curves) 
1. 160: (of surfaces) 1. 170: (of numbers^ 
in Nicomachus and lamblichus a sub- 
division otoddw. 286 

evtrtvTaa&at, construct.: special connotation 
l. 259, 289: with brrtn u 289: contrasted 
with Topal&XtaLF {apply) I. 343 : oO ffiwra- 
fcjtffcrm, (rivrdfi^rtu'rat, ll there cannot be 
constructed n l. 2119, U. 53 

ffiffT/ftn^, N^MHVNA (of ratios] II, 135, 189- 
90 : atryKeifAtPa and Owpe&ivTtL [com* 
pontndo and separando) used relatively to 
one another ll. i6S f 170 

ffiJcmjMa povdiw, B collection of units " (def. 
of number) & 280 

<7wrptfiaT(*df, collective II- 279 

(ToW/m* sphere in. 269 

cr^tuptxfe, spherical (of a particular species of 
cube number) It. 291 

it, pr^-i :■■.><: or a<fnjrl<FK<n> of solid number with 
all three sides unequal (^scalene) ll. 290 



5*6 GENERAL INDEX OF GREEK WORDS AND FORMS 



fXfoa, "relation 9; rota f^lnr, "a sort of 
relation" (in def, of ratio) n. 116-7 

trxTjffAToypaipebt y tfxTjii.tiToypa^aj representing 
(numbers) by figures of like shape r. 350 

trxyiftaTOTTGiovira or c"X*7M irwowra, '* forming 
a figure " (of a line or curve) 1. 160-1 

rafrdyn}*ip, of square number (Nicomachus) 

tb^tAttji XA^yu*, M sameness of ratios If. 1 19 
rAfioj, perfect (of a class of numbers) Ii. 

193-4 
T*ra7^PoJ t " ordered" : wrOF&mf TrptySXijjHi, 

"ordered" problem i, ra8: liiwjjlll | 

dvaXcryAi, "ordered" proportion H* 137 
TtTapayt**-**! (baXo^^j perturbed proportion 

II. 136 
fWfjMfHHtfjftdf, squaring, definitions of, 1. 149- 

T€Tp&ymra* t square : sometimes (but not in 

Euclid) any four-angled figure I, 188 
TeTpdrKtvpof, quadrilateral U 187: not a 

"polygon" EL 139 
t,u^q kOa-Xdu, segment of a circle : Tfi^fiarox 

-yw^ta, angle */a segment ft, 4: ^p r^/iari 

ywia„ angle r"/j a segment II, 4 
ro>t«)i (flforXou}, sector (of ft circle): okvtoto- 

ju^ir fWHtfti "shoemakers knife" 11. 5 
njpMf) section, = point of section [■ 170, 171, 

178: Kaurij 7-o.tijj. "common section" lit. 

rc^ion^t (of figure), seetor+iiAe II. 5 
Twurie fathpqpji, locus-theorem I, 319 
tAiw, Ikski J, 339-31 \ =room or space 
1. 13 ». : place (where things may be 
found) ^ thus r&wot &v<t\v6}uvtt, Treasury 
of Analysis I- 8, to, rapd&ofo* r6vot, 
Treasury of Paradoxes, t. 319 



rdptoi, instrument for drawing a circle I, 37 1 
TMBfTairWwi', *' the same multiple " if- 146 
Tplywo*, triangle | r& rpirXofr, t& Ji" dXXijj-- 
Xuv, triple, interwoven triangle, = penta- 
gram ti. 99 
TpiwXdattn, triple, TpnrX.otrtuv t triplicate (of 

ratios) 11* 133 
rpirktvpov, three -aided figure I. 187 
TirfxirwtWj happen : jvxbv j^ptiov, any point 
at random U 351: rv^ovsa ywla. *' any 
angle 1 ' II. 111 : AXXo, a frvx** » tocuttj toX- 
XarXiffta, "other, chance, equimultiple*" 

ftffwftft^ txeeeding, with reference to method 

of application of areas I. 36, 343-5* 

386-7 
foreprfXifr* or inttprikim, "over- perfect (of 

a class of numbers) El. 193-4 
vw&, in expressions for an angle [if M BAT 

yufta) t. 149, and a rectangle 1. 370 
vTa&irXa&ios, svo-dupiieate t = half (Nioo- 

machus) ][. l8o 
vwantfictot, laid down or assumed : t6 br* 

Ktiutvov iriwtSm, the plane of reference 

in. 974 

VT&Xtt7CU, "lS bj hypOtheStS '* I. 303, Jtl 

i>Toro\Ka.r\affitti t submultiple (Nicomachui) 

li. 180 
uTQTtfrt'jr, subtend, with ace or uw& and arc. 

U *49* ■8ft 350 
tyor, height ii. 189 

XHtptor, area IT* 1 54 

u}f.uT^n) ypau^t-ff, determinate line (cun't), 
"forming a figure" [. 160 






GENERAL INDEX. 



[The references are to volumes and pages.] 



al- 1 Abbas b. Salt! al-Jaubart I. 85 
"Abthiniathus" (or " Anlhisathus ") I. 103 
Abu '1 'Alibis aJ-Fadl b. Hitim, see an- 

Nairlii 
Abu 'Abdallih Muh. b. Mu'idh aljayyani 

I. 00 
Abu All al-Basri ]. 88 
Abu 'All al- Hasan b.al- Hasan b. al-Haitham 

[. 88, 89 
Abu Da'ud Sulaiman b. 'Uqba 1. 85, 90 
Abu Jn'far al-Khizin 1. 77, 8j 
Abu Ja'far Muh. b. Muh, b, al- Hasan 

Nasiraddin at-Tusi, set Na$iraddin 
Abu Muh. b. AbdalbaqI al-Bagdadl al-Farad] 

I. 8 it., 90 
Abu Muh. a] -I lasan b. 'Ubaidallah I). Sulai- 
man b.' Wahb i. 87 
Abu Nasr Gars al-Na'ma 1. 90 
Abu Nasr Mansur b. 'All b. 'Iraq 1. 90 
Abu Nasr Muh. b. Muh* b Tarkban b. 

Uxlag al-Farihl 1. 88 
Abu S.ihl Wljan b. Rustam al-Kuhl I. 88 
Abu Said Sinan b. Thabit b. Ourr a. ]. 88 
Abu '(Jthman ad-Dimashql 1. 15, 77 
AbO. '1 Waft al-Buzjanl I. 77, 8$,' 86 
Abu Yusuf Ya'qub b. Ishaq b. as-fiabbah al- 

Kjnd! I. 86 
Abu Yusuf Ya'qub b. Muh. ar-Razt t. 86 
Adjacent (iipt&t), meaning j. 181 
Adraslus It. 293 

Aenaeas (or Aigeias) of Hierapolis :. 18, 311 
Aganis I. 17-8, 191 

Ahmad b. al-Husain al-Ahwazt al-Katib ]- 89 
Ahmad b. 'Umai al-Karabls! 1. 8s 
al-Ahwizl 1. tit) 

Aigeias (? Aenaeas) of Hierapolis 1. 1 8, 311 
Alcinous n. 96 

Alexander Aphrodisiensis ]. -a., :o, 11. 110 
Algebra, geometrical I- 372-4 : classical 
method was that of Euct. it. {cf. Apol- 
lonius) I. 373 : preferable to semi-alge- 
braical method 1. 377-8: semi -algebraical 
method due to Heron 1* 373, and favoured 
by Pappus 1- 373 : geometrical equivalents 
of algebraical operations 1.374: algebraical 
equivalent! of propositions in Book [I-, t, 
371-3: equivalents in Book X. of pro- 
positions in algebra, ,/* - ,/X cannot be 



equal to*", m. 58-60: \fa±^i=x± t jy, 
then a=i, i=y, in. 93-4, 167-8 

All b. Ahmad Abu '1 Qasim al-Antaki I. 86 

Allman, G. J. I. 135*., 318, 351, III. t8- 
Sh439 

Alternate: (of angles) I. 308: (of ratios), 
alternately 11. 134 

Alternative proofs, interpolated I. 58, 59 : 
cf. 111. 9 and following 11. 11 : that in 
ill. to claimed by Heron 11. 13-4 

Amaldi, Ugo I. (75, 179-80, 193, aoi, 313, 
318, II. 30, 126 

Ambiguous case 1. 306-7 ; in vi. 7, II. 30B-9 

Amphinomus 1. n; h 118, t$ort. 

Amyclas of Heraclea I. 117 

Analysts {and synthesis) 1. iS ; definitions 
of, interpolated, 1. 138, ill. 441 : described 
by Pappus I. 138-9: mystery of Greek 
analysis III. 146: modern studies of Greek 
analysis 1. 139: theoretical and problem- 
atical Analysis I. 138: Treasury of Analy- 
sis {rbrot draAuo^pot) I. 8, to, Hi 138: 
method of analysis and precautions neces- 
sary to, I. 139-40: analysis and synthesis 
of problems]. J40-3: two parts of analysis 
(a) transformation, (b) resolution, and two 
parts of synthesis, (a) construction, (b) 
demonstration 1. 141 : example from 
Pappus I. 141-1: analysis should also 
reveal iiopmuit (conditions of possibility) 
1. 141 : interpolated alternative proofs of 
xiii. 1-5 by analysis and synthesis 1. 137, 
HI. 441-3 

Analytical method 1. 36 : supposed discovery 
of, by Plato 1. 134, 137 

Anaximander I. 370, II. Ill 

Anaximenes ti. 1 1 1 

Anchor- ring 1. 163 

Andron 1. 116 

Angle: curvi lineal and rectilineal, Euclid's 
definition of, t- 176 so. : definition criti* 
cised by Syrianns I. 176: Aristotle's notion 
of angle as *Ad<riF 1. 176: Apollonius' view 
of, as contraction I. 176, 177: Plutarch and 
Carpus on, I. 177: to which category does 
it belong? quantum, Plutarch, Carpus, 
"Aganis" I. 177, Euclid 1. 178; quote, 
Aristotle and Eudemus t, 1 77-8 ; relation. 



s*« 



GENERAL INDEX 



Euclid t, 17S ; Syrianus* compromise 
I* 178: treatise on the Angle by Eudemus 
I* 54' 3^i 177-8: classification of angles 
(Geminus) 1. 178-9: curvilineal and 
"miited" angles I. 16, 178-9, horn-tike 
(iccparoA&fc) i, 1 77, [78, 181, 16$, IT. 4, 
3,9, 40, tune-like {firjwo&Sfy) I. 16, 17&-0,, 
scraper-like {£wfTptxttiJjj) 1. 178: angle of & 
segment 1. 353 , 11. 4: angle of a semi- 
circle I. 181, 153, ii. 4: controversies aboui 
" angle of semicircle 17 and hornlike angle 
11. 39-43: definitions of angle classified 
1. 179: recent Italian -views 1, 179-81: 
angle as cluster of straight lines or .rays 

I. 180-1, defined by Veronese ], 180: as 
part of a plane ("angular sector* 1 ) I. 1 70- 
80: flat angle {Veronese etc.) I. 180-j, 
i6$t three kinds of angles, which is prior 
{Aristotle)? I. 181-1 : angles not less than 
two right angles not recognised as angles 
(of. Heron, Proclus, Zenodorus) (1. 47-9: 
did Euclid extend "angle" to angles 
greater than two right angles in vi. 33? 

II. 175-6: adjacent angles 1- 181 : alternate 
I. 308; similar [ = equal) I. 178, 1 81, a£l 3 

vertical J. 178: exterior and interior 
(to a figure) 1. 163, 180: exterior when 
re-entrant I. 1163, in which case we have a 
hollow-angled figure I. 17, 188, If. 48: 
interior and opposite 1. 1S0; construction 
by Apollonius of angle equal to angle 
1. *o6; angle in a semicircle, theorem of, 
k V7" 1 )' trisection of angle, by con. 
choid of Kicomedes 3, if\=,-fi, by quad rat rix 
of Hippias u *66, by spiral of Archimedes 
I. 167: dihedral angle ill. 164-5; solid 
angle ill. 161, ■367-8 
Annex {rpooapfid^ovoa) = the straight line 
which, when added to a compound ir- 
rational straight Line formed by subtraction, 
makes up the greater "term," i.e. the 
negative "term^ ill. T59 
al-Antikl 1. 86 
Antecedent* (leading terms in proportion) 11. 

'34 
* ( Anthisathus"(or " Abthiniathns ") 1. 303 
Antiparallels ; may be used for construction 

of vi. |«| if, 115 
Antiphon l. 7 n- t 35 

Apastamba-Sulba-Sutra 1. 35a: evidence in, 
as to early discovery of End- 1. 47 and use 
of gnomon u 360-4: Bilrk's claim that 
Indians had discovered the irrational 1. 
363-4: approximation to ^1 and Thibaut's 
explanation 1. 361, 363-4: inaccurate 
values of ■* in, j. 364 
Apollodorus u Logisticus" j. 37, 319, 351 
Apollonius : disparaged by Pappus in com- 
parison with Euclid 1. 3: supposed by 
some Arabians to be author of the Ele- 
ments I. $: a "carpenter" I. 5: on ele- 
mentary geometry 1. 41; on the line 1. 
1*9: on the angle I, 176: general defini- 
tion of diameter l. 335: tried to prove 
axioms 1, 4*, 6a, 114-3: his "general 



treatise" l. 42: constructions by, foi 
bisection of straight Line 1. 168, for * 
perpendicular I 170, for an angle equal to 
an angle I. 396: on parallel -axiom (?) 
I- 43-3 1 adaptation to conies of theory of 
application of areas r» 344-5: geometrical 
algebra in, t, 373: Plane foci^ I. l 4, ■259,330-, 
theorem from (arising out of Eucl- VI. 3)1 
also found in Aristotle It, 198-300: Plane 
r*i«r«* c 151, problem from, 11. 8i, lemma 
by Pappus on, if, 64-5 : comparison of do- 
decahedron and icosahedrou U 6, 111. 419, 
512, 513: on the rochliai 1. 34, 41, t&i: 
on w unordered" irrationals I- 41,715, III. 
3, 10, 1^6, 255-9: general definition of ob- 
lique (circular) cone lit* 270: I- 138, 1&8, 
111, 111, 146, 359> 37°. S73* »- 75* 19A 

158, III. 164, 267 

Apeiowc; compound irrational straight line 
(difference between two "terms* 1 ) ill. 7: 
defined tit. 158-9: connected by Theae- 
tetus with harmonic mean ill. 3, 4: 
biquadratic from which it arises III- 7: 
uniquely formed 111. i6?S r . first, second, 
thirds fourth, fifth and sixth apotomes, 

auadralicG from which arising in. 5-6, 
efined ill. 177, and found respectively 
(x. 85-90) in. 1 78-90; apotome equivalent 
to square root 01 first apotome ill* 190-4: 
first, jetondt third, fourth, fifth and sixth 
apotomes equivalent to squares of apotome, 
first apotome of a medial etc. III. 313-19: 
apotome cannot be iwmifl/ also HI. 140-1 : 
different from medial [straight line) and 
from other irrationals of same series with 
itself 111. 241 : used 10 rationalise binomial 
with proportional terms til. 143-81 '5J-4 
Apotome of a medial (straight line); first and 
second, and biquadratics of which they are 
roots in* 7: first apotome of a medial 
defined HI, 159-60, uniquely formed ill. 
168-9, equivalent to square root of second 
apotome in. 104-8: second apotome of a 
medial, defined in. t6i-i, uniquely formed 
ni. 170-1, equivalent to square root of 
third apotome III. 199 -101 
Application of areas 1. 36, 343-5- contrasted 
with exceeding and falling-short I* 343: 
complete method equivalent to geometrical 
solution of mixed quadratic equation 1- 
3 44-5 > 3*3-5. 386-8, 11. 187, 158-60, 
W3"* 166-7 * adaptation to conies (Apol- 
lonius) h 344-5: application contrasted 
with construction (Proclus) l« 343 
Approximations ; 7/5 as approximation to v /i 
(Pythagoreans and Plato) 11. 1 19 : approxi- 
mations to J$ in Archimedes and (in 
sexagesimal fractions) in Ptolemy 11, 119: 
to ir (Archimedes) n, 119: to V4500 
(Theon of Alexandria) P. no: remarkably 
close approximations {staled in sexagesimal 
fractions) in scholia to Book X. t l< 74 n. 
H Aqaton '* [, 88 

Arabian editors and commentators 1, 75- 
0° 



GENERAL INDEX 



5*9 



Arabic numerals in scholia to Book X., 
i xth c.j i* 71 

Archibald) ]■:. C. 1. 9 »., ro 

Archimedes: "postulates" in, 1. no, liA 4 , 
"poris-ms** in, I. liw,, 1 j : on straight 
lint 1. 166: on plane 1, 171-1: Liber 
aszumptarum , proposition from, It. 65 : 
approximations to ^3, square roots of large 
numbers and to t, ii. 119: extension of 
a proportion between commensurable to 
coverincommensurablesll. 103: " Axiom '* 
of (called however " lemma, assumption, 
by A. himself) I. 334 : relation of "Axiom " 
10 X* t, ill. t$-6 1 ** Axiom" already 
used by Eudoxus and mentioned by 
Aristotle ill. 16: proved by means of 
Dedekind's Postulate (Stolz) ill. 16 : on 
discovery by Eudoxus of method of ex- 
haustion 111. 365-6, 37* : °e w fragment 
of, * Method (tyofotj of Archimedes about 
mechanics] theorems," or c-<p&5tov, dis- 
covered by Heiberg and published by 
him ll* 40, tn. 366-8, adds new chapter to 
history of integral calculus, which the 
method actually is, tit, 366-7 : application 
to area of parabolic segment, ibid- : spiral 
of Archimedes I. 16, 167: l. 10, 21, 116, 
141, 225, ?49' 370i Hi ■ 3<5 . 190, ill. 346, 
*70, 375 

Archytas t, jo : proof that the